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Chapter 1
Passage of Charged Particles Through Matter
1.1 Various Types of Processes
When charged particles pass through matter, the following processes may take
place:
(1) Inelastic collisions with the bound electrons of the atoms of the medium, in
which case the particle energy is spent in the excitation or ionization of atoms
and molecules. The energy losses of this kind of collisions are called ionization losses (collision losses) to distinguish them from radiation losses that are
concerned with the generation of bremsstrahlung.
(2) Inelastic collisions with nuclei, leading to the production of bremsstrahlung
quanta, to the excitation of nuclear levels, or to the nuclear reactions.
(3) Elastic collisions with nuclei, in which part of the kinetic energy of the incident particle is transferred to the recoil nuclei. However, the total kinetic energy of the colliding particles remains unchanged. A particular type of elastic
scattering is the Rutherford scattering which results from the interaction of a
charged particle with the Coulomb field of the target nucleus in single encounters. When thick materials are used, cumulative single scatterings give rise to
the phenomenon of multiple scattering.
(4) Elastic collisions with bound electrons.
(5) Cerenkov effect, i.e. emission of light by charged particles passing through matter with a velocity exceeding the velocity of light waves in the given medium.
1.2 Kinematics
1.2.1 Laboratory (Lab) System (LS) and Centre of Mass
System (CM)
In order to describe the motion of particles in the collision problem one must choose
a definite frame of reference (co-ordinate system). Two frames of reference are imA. Kamal, Nuclear Physics, Graduate Texts in Physics,
DOI 10.1007/978-3-642-38655-8_1, © Springer-Verlag Berlin Heidelberg 2014
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Passage of Charged Particles Through Matter
Fig. 1.1 The position vectors
m1 and m2 and their centre of
mass is shown
portant, one is the lab system (LS) and the other one is centre of mass system (CMS).
In the lab system, the observer who is at rest in the lab views the collision process.
In the CM system the centre of mass is at rest initially and always. Observations
are usually made in the lab system but theoretical calculations are made in the CM
system. It is of great interest to find out how various quantities like velocity, angle
of scattering, etc. are related in these two systems. It is easier to perform calculations in the CM system rather than in the lab system. For, the great merit of CM
system is that the total linear momentum of particles is always zero so that in the
two-body process particles move directly towards each other before the collision
and they recede in the opposite direction after the collision.
The collision process in the CM system may be visualized as the one in which
a particle of reduced mass μ = m1 m2 /(m1 + m2 ) moving with initial velocity u1
collides with a fixed scattering centre. Here, u1 is the initial velocity of m1 moving
towards the target particle of mass m2 at rest.
1.2.2 Total Linear Momentum in the CM System Is Zero
In Fig. 1.1, the position of the centre of mass of two particles m1 and m2 is shown
by C. The position of masses m1 and m2 are indicated by the position vectors r1
and r2 and that of the centre of mass by R. By definition
m1 r1 + m2 r2
M
MR = m1 r1 + m2 r2
R=
or
Differentiating with respect to time
M Ṙ = m1 ṙ + m2 ṙ
or
Mvc = m1 u1 + m2 u2
where u1 and u2 are the initial velocities of particles 1 and 2, respectively and vc is
the CM velocity. Since m2 is initially at rest, u2 = 0, and the centre of mass which
is located at M = m1 + m2 , must move in the lab system towards m2 with velocity
vc =
m1 u1
m1 + m2
(1.1)
1.2 Kinematics
3
Fig. 1.2 Collision in the LS and CMS are shown
In the lab system, let m1 move from left to right with initial velocity u1 , m2 being
initially at rest as in Fig. 1.2. As m2 is initially at rest, its initial velocity in the
CMS must be just equal to vc in magnitude but oppositely directed. Denoting the
velocities in the CMS by asterisk (∗ ) we get
u∗2 = vc =
m1 u1
m1 + m2
u∗2 = −vc
(1.2)
(1.3)
The initial velocity of m1 in the CMS is reduced by an amount equal to vc
u∗1 = u1 − vc
m1 u1
m2 u1
u∗1 = u1 −
=
m1 + m2 m1 + m2
where we have used (1.2). Total initial linear momentum of m1 and m2 in the CMS is
P ∗ = P1∗ + P2∗ = m1 u∗1 + m1 u∗2 =
m1 m2 u∗1
m2 m1 u∗1
−
=0
m1 + m2 m1 + m2
(1.4)
where we have used (1.2), (1.3) and (1.4). Thus total linear momentum of particles
in the CMS is zero before the collision and by conservation of momentum, this must
be so after the collision.
1.2.3 Relation Between Velocities in the LS and CMS
Lab system
m1 : u1 ,
m2 : u2 = 0,
CM system
m2 u1
m1 + m2
m1 u1
u∗2 =
m1 + m2
u∗1 =
(1.5)
(1.6)
For elastic collisions, both momentum and kinetic energy must be conserved. This
implies that the respective velocities of the particles before and after the collisions
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in the CMS must be equal
u∗2 = v2∗
u∗1 = v1∗ ;
m2 u1
v1∗ =
m1 + m2
(1.7)
(1.8)
Observe that in both the LS and CMS, the relative velocity of the two particles is
equal to u1 . We know
u(rel) = u∗1 + u∗2 =
m2 u1
m1 u1
+
= u1
m1 + m2 m1 + m2
Using (1.2) and (1.8),
vc
m1
=
= γ.
∗
v1
m2
(1.9)
It is seen that if m1 < m2 , then vc < v1∗ and if m1 > m2 , vc > v1∗ .
1.2.4 Relation Between the Angles in LS and CMS
Figure 1.3 shows the scattering and recoil angles in the LS and CMS.
The lab velocity v1 of m1 after the collision is obtained by combining vectorially
its velocity v1∗ in the CMS and the CM velocity vc (Fig. 1.4)
v1 = v1∗ + vc
Let m1 be scattered at an angle θ as seen in the LS, its corresponding angle in the
CMS being θ ∗ . In the velocity triangle (Fig. 1.4) resolving the velocities along the
x-axis and y-axis, we get
v1 sin θ = v1∗ sin θ ∗
v1 cos θ =
v1∗ cos θ ∗
(1.10)
+ vc
(1.11)
Dividing (1.10) by (1.11)
tan θ =
v1∗ sin θ ∗
∗
v1 cos θ ∗ + vc
=
sin θ ∗
sin θ ∗
=
∗
cos θ ∗ + vc /v1
cos θ ∗ + m1 /m2
where we have used (1.9).
Special cases
(i) m1 m2 ; θ θ ∗ . Here vc → 0 and the CMS is reduced to the LS.
Example: α-gold nucleus scattering.
(ii) m1 m2 ; θ 0◦ .
Example: nucleus-electron
scattering.
∗
(iii) m1 = m2 ; tan θ = cossinθ ∗θ+1 = tan 12 θ ∗ so that θ = 12 θ ∗ .
Example: proton-proton scattering.
(1.12)
1.2 Kinematics
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Fig. 1.3 Relation between the angles in LS and CMS
Fig. 1.4 Velocity triangle for
the scattered particle
1.2.5 Recoil Angle
Let m2 recoil with velocity v2 at an angle φ with the incident direction in the LS.
Let its velocity be v2∗ at angle φ ∗ in the CMS. From the velocity triangle in Fig. 1.5,
we get
v2 sin φ = v2∗ sin φ ∗
(1.13)
v2 cos φ = v2∗ cos φ ∗ + vc
(1.14)
Dividing (1.13) by (1.14)
tan φ =
v2∗ sin φ ∗
v2∗ cos φ ∗ + vc
but by (1.1), (1.6) and (1.7), v2∗ = vc
∴
tan φ =
φ∗
sin φ ∗
= tan
∗
cos φ + 1
2
or
φ = φ ∗ /2 (regardless of the ratio m1 /m2 )
(1.15)
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Fig. 1.5 Velocity triangle for
the recoil particle
1.2.6 Limits on the Scattering Angle θ
Case (i) m2 > m1 , or γ < 1; i.e. v1∗ > vc .
In Fig. 1.6, the circle is drawn with O as the centre and radius OP = v1∗ . A is
a point within the circle such that AO = vc , and the line AOB represents the incident direction. As before, the lab velocity of m1 is v1 which is obtained by compounding v1∗ and vc vectorially. The lab angle θ = angle P AO and the CM angle
θ ∗ = angle P OB. As the point P moves counterclockwise on the circumference,
θ ∗ increases and so does θ . When P approaches P , θ ∗ = θ = π . Thus, θ increases
monotonically from 0 to π , and in this case there is no restriction on the scattering angle in the LS. In other words, m1 can be scattered in completely backward
direction.
Case (ii) m2 = m1 , or γ = 1; i.e. v1∗ = vc .
Here A lies on the circumference of the circle (Fig. 1.7). As θ ∗ increases, θ also
increases. But when P approaches A, P A becomes tangential at A and so θ → π2 .
θ varies from 0 to π . Thus, in this case m1 can be scattered up to a maximum angle of π/2 but not beyond. In other words, backward scattering in the LS is not
permissible.
Case (iii) m2 < m1 , or γ > 1, i.e. v1∗ < vc .
Here A lies outside the circle (Fig. 1.8). There are two positions P and P l for
which the same scattering angle θ is obtained for two different values of θ ∗ . As P
moves back on the circumference, θ increases. The maximum angle θm is reached
when AP becomes tangent to the circle (Fig. 1.9). In that case
sin θm =
v ∗ m2
OP
= 1 =
AO
vc
m1
or
θm = sin−1 (m2 /m1 )
Thus, there is a limitation on the scattering angle when m2 < m1 . θ first increases
from 0 to a maximum value sin−1 (1/γ ) which is less than π/2, as θ ∗ increases from
0 to cos−1 (−1/γ ). θ then decreases to 0 as θ ∗ further increases to π . At a given
angle θ between 0 and sin−1 (1/γ ), there will be two groups of particles associated
with different velocities corresponding to the two values of θ ∗ .
1.2 Kinematics
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Fig. 1.6 Limits on the
scattering angle for m2 > m1 ,
θ max = π
Fig. 1.7 Limits on the
scattering angle θ for
m1 = m2 , θ max = π/2
Fig. 1.8 Limits on the
scattering angle θ . For
m2 < m1 ,
2
θ(max) = sin−1 ( m
m1 )
Fig. 1.9 θmax is reached
when AP is a tangent to the
circle
1.2.7 Limits on the Recoil Angle φ
Since v2∗ = vc (always), φ = 12 φ ∗ by (1.15). Since the maximum angle of φ ∗ is π ,
the maximum angle φm is 12 π . In other words, the target particle cannot recoil in the
backward hemisphere.
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Fig. 1.10 Azimuth angle β is
measured with respect to the
positive axis in the xy plane
⊥ to the direction of
incidence
Fig. 1.11 Element of solid
angle dΩ = ds/r 2
1.2.8 Scattering in Three Dimensions
Since scattering is described under central forces, a particle which is incident on a
target particle and initially moves in a certain plane would be necessarily confined
to this plane after the scattering because of the conservation of angular momentum.
Thus, a single scattering event is completely described in two dimensions. However,
in practice one is concerned with a flux of particles incident say along the z-axis on
a target material. Since various particles proceed in different planes, the scattering
on the whole will be in three dimensions. In order to fix the orientation of the plane
of scattering, we need to introduce the azimuth angle β which is measured with
respect to the positive x-axis in the xy plane, Fig. 1.10. We must also consider the
element of solid angle into which the particles are scattered. This is illustrated in
Fig. 1.11.
Let dA denote an element of surface area and connect all points on the boundary
of dA to O so as to form a cone. Let ds be the area of that portion of a sphere
with O as the centre and radius r which is cut out by this cone. The solid angle
subtended by dA at O is defined as dΩ = ds/r 2 and is numerically equal to the
area cut out by a sphere with centre O and unit radius. From Fig. 1.12, it is seen that
ds = r 2 sin θ dθ dβ so that dΩ = sin θ dθ dβ = 2π sin θ dθ , where we have integrated
over dβ. When the scattering is independent of the azimuth angle then the area
subtended at O is due to the entire circular strip, ds = 2πr 2 sin θ dθ as in Fig. 1.13,
so that the element of solid angle dΩ = 2π sin θ dθ . Observe that the maximum
solid angle is 4π since it is given by the entire surface area of a sphere (4πr 2 )
divided by r 2 .
1.2 Kinematics
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Fig. 1.12 Elements of solid
angle for a general case
Fig. 1.13 Element of solid
angle for an azimuthal
symmetry
1.2.9 Scattering Cross-Section
In order to describe the angular distribution of particles scattered by target particles
which are initially stationary, the concept of cross-section is introduced. Let a uniform parallel flux of N0 particles be incident per unit area normal to the direction
per unit time on a group of n scattering centres. Let N particles be scattered per unit
time into a small solid angle dΩ centred towards a direction which has polar angle
θ and azimuth angle β with respect to the incident direction as polar axis. N will
be proportional to N0 , n and dΩ provided the flux is small enough to ensure that
the incident particles do not interfere with one another, that there is no appreciable
decrease in the number of scattering centres on account of their being knocked out
due to collisions and that the incident particles are far enough apart so that each
collision is made only by one of them.
The number of incident particles that emerge per unit time in dΩ can be written
as:
N = nN0 σ (θ, β)dΩ
(1.16)
where the proportionality factor σ (θ, β) is called the differential scattering crosssection. The quantity σ (θ, β) is a measure of the probability of scattering in a given
direction (θ, β), per unit solid angle from the given nucleus.
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The integral of σ (θ, β) over the sphere is called the total scattering cross-section
σ = σ (θ, β)dΩ
(1.17)
σ has the dimension of area. The unit of σ is a Barn (b). 1 b = 10−24 cm2 . The unit
of σ (θ, β) is Barn/Steradian, where Steradian (sr) is the unit of solid angle. σ (θ, β)
is also written as dσ (θ, β)/dΩ.
Compared to the scattering in two dimensions the only additional parameter
which has been introduced to describe scattering in three dimensions is known as
the azimuth angle β.
1.2.10 Relation Between Differential Scattering Cross-Sections
The relation between the differential cross-section in the laboratory and the centreof-mass co-ordinate systems can be obtained from their definition which implies that
the number of particles scattered into the element of solid angle dΩ about θ , β is the
same as are scattered into dΩ ∗ about θ ∗ , β ∗ . In polar co-ordinates dΩ = sin θ dθ dβ
and dΩ ∗ = sin∗ dθ ∗ dβ ∗
σ (θ, β) sin θ dθ dβ = σ θ ∗ , β ∗ sin θ ∗ dθ ∗ dβ ∗
but
β = β∗
∴
sin∗ dθ ∗
σ (θ ) = σ θ ∗
sin θ dθ
(1.18)
Differentiating (1.12)
sec2 θ dθ =
|1 + γ cos θ ∗ |dθ ∗
(cos θ ∗ + γ )2
(1.19)
Using (1.12) and the identity, sec2 θ = 1 + tan2 θ (1.19) is easily reduced to the
form:
dθ ∗ [1 + 2γ cos θ ∗ + γ 2 ]
=
dθ
|1 + γ cos θ ∗ |
(1.20)
sin θ ∗
sin θ
=
cos θ
cos θ ∗ + γ
(1.12)
Also
tan θ =
whence
sin θ ∗ cos θ ∗ + γ
=
sin θ
cos θ
(1.21)
1.2 Kinematics
11
Also
1
= sec θ = 1 + tan2 θ
cos θ
Using (1.12) in (1.22) and re-arranging them we get
cos θ ∗ + γ
=
cos θ
1 + 2γ cos θ ∗ + γ 2 =
(1.22)
sin θ ∗
sin θ
(1.23)
Using (1.20) and (1.23) in (1.18)
σ (θ ) =
(1 + γ 2 + 2γ cos θ ∗ )3/2 σ (θ ∗ )
|1 + γ cos θ ∗ |
(1.24)
It must be pointed out that the total cross-section is the same for both lab and CM
systems, since the occurrence of total number of collisions is independent of the
mode of description of the process.
1.2.11 Kinematics of Elastic Collisions
We have to obtain an expression for velocity v1 as a function of scattering angle θ .
From the velocity triangle (Fig. 1.14)
v1∗2 = vc2 − 2v1 vc cos θ + v12
(1.25)
Substituting for vc and v1∗ from (1.1) and (1.8), (1.25) becomes
v12 −
(m1 − m2 )
2m1 u1 cos θ v1
+ u21
=0
m1 + m2
m1 + m2
This is a quadratic equation in v1 whose solutions are found to be
m22
m1 u1
2
cos θ ±
v1 =
− sin θ
m1 + m2
m21
(1.26)
For the special case, m1 = m2 , (1.26) simplifies to:
v1 = u1 cos θ
so that the ratio of kinetic energy T1 and T0 of the scattered and incident particle
becomes
v2
T1
= 12 = cos2 θ
T0 u1
with the restriction, θ ≤ 90◦ , as pointed out earlier.
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Fig. 1.14 Velocity triangle
for the scattered particle
Fig. 1.15 Velocity triangle
for recoil particle
1.2.12 To Derive an Expression for the Recoil Velocity v2
as a Function of φ
From the velocity triangle (Fig. 1.15)
v2∗2 = vc2 + v22 − 2vc v2 cos φ
Since
v2∗ = vc
v2 = 2vc cos φ =
2m1 u1
cos φ
m1 + m2
(1.27)
where we have used (1.1).
The ratio of kinetic energy of the recoil particle and original kinetic energy of the
incident particle is:
T2 m2 v22
4m1 m2
=
=
cos2 φ
2
T0 m1 u1 (m1 + m2 )2
(1.28)
For the special case m1 = m2
T2
= cos2 φ
T0
(1.29)
1.2.13 Available Energy in the Lab System and CM System
Assuming that the target particle is at rest before the collision, total kinetic energy
in the lab system is
T = T0 ,
1
with T0 = m1 u21
2
(1.30)
1.2 Kinematics
13
In the CM system, m1 has kinetic energy
m2 u1 2
1 2 1
T1∗ = m1 u∗1 = m1
2
2
m1 + m2
where we have used (1.4).
In the CM system, m2 has kinetic energy:
T2∗
m1 u1 2
1 ∗ 2 1
= m2 u2 = m2
2
2
m1 + m2
where we have used (1.2).
Total kinetic energy available in the CM system is:
T ∗ = T1∗ + T2∗ =
1 m1 m2 2 1 2
u = μu
2 m1 + m2 1 2 1
(1.31)
where μ is the reduced mass.
Formula (1.31) shows that the two-body problem is reduced to a one-body problem by imagining that a particle of mass μ = m1 m2 /(m1 + m2 ) is directed towards
a scattering centre, with the velocity u1 . Using (1.30) in (1.31)
T∗ =
m2 T0
m1 + m2
where T ∗ < T0 .
Thus less energy of motion is available in the CM system. It can easily be shown
that the difference in energy in the lab and CM systems is associated with the motion
of CM system
1
1
1 m1 m2 2 1 m21 u21
T = T0 − T ∗ = m1 u21 −
= (m1 + m2 )vc2
u =
2
2 m1 + m2 1 2 (m1 + m2 ) 2
(1.32)
where we have used (1.1).
Formula (1.32) shows that the difference of energy goes into the motion of CM
of mass (m1 + m2 ) with velocity vc . We conclude that in the CM system energy that
is available is always less than that in the lab system, for some energy must go into
the motion of CM system.
For the special case m1 = m2
1
1
T ∗ = m1 u21 = T0
4
2
This fact has a bearing on production thresholds, i.e. minimum energy that is to
be provided in order to produce particles. Consider, for example, the case of pion
production in proton-proton collisions. The rest mass of pion is only 140 MeV/c2 .
However, this much energy must be available in the CM system. This means that in
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the lab system, the incident proton must have double this energy viz, 280 MeV in
order to produce a pion. Relativistic calculations actually give a value of 290 MeV.
These considerations are also important in the invention of a new class of high
energy accelerators in recent years, in which colliding beams of particles are used;
i.e. one beam travels in one direction and is intercepted by another beam of similar
or dissimilar particles of the same energy moving in the opposite direction. In this
case, the CM system is realized in the laboratory itself and lot of energy is made
available.
Example 1.1 If a particle of mass m collides elastically with one of mass M at rest,
and if the former is scattered at an angle θ and the latter recoils at an angle φ with
respect to the line of motion of the incident particle, then show that
tan θ =
m
M
sin 2φ
− cos 2φ
Hence, show that
sin(2φ + θ )
m
=
M
sin θ
Solution
tan θ =
∴
sin θ ∗
cos θ ∗ + m/M
but θ ∗ = π − φ ∗ = π − 2φ
sin θ ∗ = sin(π − 2φ) = sin 2φ
cos θ ∗ = cos(π − 2φ) = − cos 2φ
∴
tan θ =
sin 2φ
sin θ
=
cos θ
m/M − cos 2φ
Re-arranging the above we get
m
sin θ = sin θ cos 2φ + cos θ sin 2φ = sin(θ + 2φ)
M
sin(2φ + θ )
m/M =
sin θ
Example 1.2 A particle makes an elastic collision with another particle of identical
mass, initially at rest. Prove that after scattering, the lab angle between the outgoing
particles is 90°.
Solution
First Method
We use the lab system. Let the particle of mass m, momentum P and kinetic energy
T move along the x-axis. After collision the particles have momenta P1 and P2 at
1.2 Kinematics
15
Fig. 1.16 Elastic collision in
LS for m1 = m2
angles θ and φ as in Fig. 1.16. Conservation of momentum along the direction of
incidence (x-axis) gives
P = P1 cos θ + P2 cos φ
(1)
Conservation of momentum along the perpendicular direction (y-axis) yields
P2 sin φ − P1 sin θ = 0 or
0 = P1 sin θ − P2 sin φ
(2)
Squaring and adding (1) and (2) and simplifying we get
P 2 = P12 + P22 + 2P1 P2 cos(θ + φ)
(3)
Energy conservation gives:
P2
P2
P2
= 1 + 2
2m 2m 2m
or
P 2 = P12 + P22
(4)
(5)
Using (5) in (3):
2P1 P2 cos(θ + φ) = 0
Since P1 = 0; P2 = 0; cos(θ + φ) = 0 or θ + φ = 90◦ .
Second Method (Vector Method)
Momentum of conservation demands that (Fig. 1.17)
P1 + P2 = P
Taking the scalar product
P · P = (P 1 + P 2 ) · (P 1 + P 2 )
P · P = P1 · P1 + P2 · P2 + P1 · P2 + P2 · P1
P 2 = P 21 + P 22 + 2P 1 · P 2
(6)
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Fig. 1.17 Momentum
triangle for the elastic
collision
since the scalar product of a vector by itself is the square of the magnitude of the
vectors and the order of scalar product is immaterial.
In view of energy conservation, i.e. with the aid of (5), we find
2P1 P2 = 0
or
2P1 P2 cos α = 0
where α is the angle between the vectors P1 and P2
∴
α = 90◦
Third Method
Because of the conservation of momentum, P1 , P2 , and P form a closed triangle.
Their magnitudes are indicated in Fig. 1.17. Because of energy conservation we
further have the relation:
P12 + P22 = P 2
i.e. the triangle must be a right angle triangle. Hence, α = 90◦ .
Fourth Method
We use the following formula for transformation of angles between LS and the
m
CMS. Set M
= 1 in (1.12)
tan θ =
∴
θ=
sin θ ∗
θ∗
= tan
∗
1 + cos θ
2
θ∗
2
But θ ∗ = π − φ ∗ and φ ∗ = 2φ, always
φ ∗ = 2θ = π − 2φ
whence θ + φ =
π
2
Example 1.3 Show that if a particle of mass m is scattered by a particle of mass
M initially at rest, then the angle between the final directions of motion in the lab
system is:
1
1 −1 m
π
+ θ − sin
sin θ
2
2
2
M
1.2 Kinematics
17
Hence, show that for particles of equal masses, the angle between final directions of
motion is always 90°.
Solution From Example 1.1 we get
sin−1
m
sin(2φ + θ )
=
or
M
sin θ
m
sin θ = sin π − (2φ + θ )
M
m
sin θ = π − (2φ + θ )
M
θ
π
1
m
φ + = − sin−1
sin θ
2
2
2
M
The angle between the final directions of motion is
α=φ+θ =
π
1
1
m
+ θ − sin−1
sin θ
2
2
2
M
For m/M = 1, α reduces to 12 π .
Example 1.4 At low energies, neutron-proton scattering is isotropic in the Csystem. If K is neutron lab energy and σ the total cross section, show that in the
lab, the proton energy distribution is
dσp /dKp = const = σ/K0
Solution In Fig. 1.18, ABC is the momentum triangle. Since the angle between
the scattered neutron and recoil proton must be a right angle
PP = P0 cos φ
KP = PP2 /2mP
and K0 = P02 /2mn
but, mp mn = m
KP /K0 = PP2 /P02 = cos2 φ
or
KP = K0 cos2 φ
dKp = −2K0 cos φ sin φdφ = −K0 sin 2φdφ
but, φ = φ ∗ /2 and dφ = dφ ∗ /2
1
dKp = − K0 sin φ ∗ dφ ∗
2
Isotropy requires that dσp /dΩ ∗ = σ/4π
dσp
dσp dΩ ∗
σ 2π sin φ ∗ dφ ∗
σ
=−
=
=
= const
∗
1
∗
∗
dKp
dΩ dKp
4π 2 K0 sin φ dφ
K0
18
1
Passage of Charged Particles Through Matter
Fig. 1.18 Momentum
triangle for n–p scattering
Negative sign is introduced in the last equation because as φ ∗ increases Kp decreases.
Example 1.5 A beam of particles of mass m is elastically scattered by target particles of mass M initially at rest. If the angular distribution is spherically symmetrical
in the centre of mass system, what is it for M in the lab system?
Solution
σ
= const
4π
sin φ ∗ dφ ∗ ∗ σ (φ) =
σ φ
sin φdφ
σ (φ ∗ ) =
but
φ ∗ = 2φ
σ (φ) =
and dφ ∗ = 2dφ
σ
sin 2φ2dφ σ
= cos φ
sin φdφ 4π
π
It may be recalled that φ is limited to 90°, i.e. the target particles can recoil only in
the forward hemisphere in the lab system. It is instructive to note that
σ (φ)dΩ =
1
2π
0
= 2σ
σ
cos φ2π sin φdφ
π
1
sin φd(sin φ) = σ
(as it should)
0
Example 1.6 Small balls of negligible radii are projected against an infinitely heavy
sphere of radius R. Assuming the balls are elastically scattered and bounce off in
such a way that the angle of reflection (r) is equal to the angle of incidence (i).
Prove that the scattering is isotropic, i.e. σ (φ) is independent of θ and that the total
cross-section is equal to πR 2 .
Solution Let b be the impact parameter (perpendicular distance of the line of flight
from the central axis). The angle of incidence and reflection are measured with
1.2 Kinematics
19
Fig. 1.19 Scattering of a
small ball off a heavy sphere
respect to the normal at the point of scattering (Fig. 1.19)
bdb
sin θ dθ
(1)
θ = π − (i + r) = π − 2i
(2)
σ (θ ) = −
See Eq. (1.56).
From the geometry of Fig. 1.19
sin θ = sin(π − 2i) = sin 2i = 2 sin i cos i
(3)
since
r = i,
and dθ = −2di
(4)
but
b = R sin i
(5)
Hence
db = R cos idi
(6)
σ (θ ) = R 2 /4.
using (2), (4), (5) and (6) in (1) and cancelling various factors; we get
The right hand side of σ (θ ) is independent of θ , the scattering angle. Hence, the
scattering is isotropic, i.e. equally in all directions.
The total cross-section is given by
π 2
R
2π sin θ dθ = πR 2
σ = σ (θ )dΩ =
4
0
Observe that the total cross-section σ has the dimension of area, and in the above
example it is equal to the projected area of the sphere. It is, therefore, called geometrical cross-section. Formula (5) shows that if b = O (head-on collision), i = 0
and from (2), θ = 180◦ . Thus, in this case the ball bounces in the opposite direction.
Again, when b = R, i = 90◦ and θ = 0◦ (glancing collision). Thus, the ball having
hit the edge of the sphere, does not suffer any deviation and continues its flight in
the incident direction. Of course, if b > R, the ball goes undeviated and there is no
scattering. The above example shows the concept of σ (θ ) and σ .
20
1
Passage of Charged Particles Through Matter
1.3 Rutherford Scattering
1.3.1 Derivation of Scattering Formula
Here we are concerned with the scattering (deflection) of point charged particles by
a massive centre of electric force. The force is assumed to be central, i.e. directed
along the line joining the centres of the colliding particles. Rutherford supposed that
all the positive charge and hence practically all the mass of the atom is concentrated
in a core or nucleus whose volume is very much less than that of the atom. Outside
the nucleus is a relatively empty space only occupied by a few electrons. Suppose,
an alpha particle is fired against the atom, then it is permitted to penetrate close to
the nucleus and owing to the electrical interaction with the nucleus it may suffer a
large angle deflection and recede from the nucleus and the effect of widely dispersed
electrons can be neglected.
A particle of charge +ze (for alpha particle, z = 2) at a distance r from the
nucleus of charge +Ze (Z being the atomic number) experiences a repulsive force
zZe2 /r 2 (Coulomb’s inverse square law) and the corresponding potential energy
will be zZe2 /r. When the incident (incoming) particle is at a very large distance,
the potential energy will be zero, and the energy is entirely kinetic due to the motion
of the particle.
Let the particle of charge +ze and mass m be incident from a very large distance from the nucleus (for example at a point A, Fig. 1.20), with velocity v0 . In
the absence of forces between the nucleus (henceforth called target nucleus) at
F and the incident particle, the particle would have continued to move along the
straight line AOB. Let F Q be perpendicular on AOB. Then b = F Q is called
impact parameter. Since the target nucleus is considered infinitely heavy, it does
not move during the encounter. The analysis will therefore be made in the lab system. The force is repulsive and central. We shall prove that under the influence of
Coulomb’s force, the trajectory is a hyperbola with the external focus F at the nucleus.
It is convenient to introduce the polar co-ordinates r, θ . The radial distance r
is measured from the focus F and the angle θ with the x-axis, which is arbitrarily
chosen. When the particle is near the nucleus, it will be deviated from the rectilinear
trajectory under the action of electrical forces and its typical position at some instant
would be at some point P with co-ordinates (r, θ ) and velocity v. Since the force
is repulsive, v < v0 . After the complete encounter, the particle is deflected through
angle θ0 and would recede to a remote distance beyond which it would continue
along the straight path OD, and at a distant point like D. It would again have the
original speed v0 , as the potential energy again approaches zero.
If the original path of the incident particle lies in a plane (here plane of paper)
then because angular momentum is conserved, the particle would continue its path
in the same plane throughout.
The conservation of energy gives:
1 2 1 2
mv = mv + zZe2 /r
2 0 2
(1.33)
1.3 Rutherford Scattering
21
Fig. 1.20 Geometry of Rutherford scattering
In polar co-ordinates, the components of velocity along and perpendicular to r are
ṙ and r θ̇ , respectively, so that
v 2 = (ṙ)2 + (r θ̇ )2
(1.34)
where dot means differentiation with respect to time. Eliminating v between (1.33)
and (1.34) and re-arranging we get:
1=
1
2zZe2
+ 2 (ṙ)2 + (r θ̇ )2
2
mv0 r
v0
(1.35)
Also in the absence of external forces, angular momentum must be conserved. Take
the angular momentum about an axis passing through the nucleus and perpendicular
to the plane. Initially the momentum mv0 is in the direction AB and the perpendicular distance F Q is b. Therefore, the initial angular momentum = (mv0 )b. At the
point P , the component of velocity perpendicular to F P is r θ̇ , and the distance
F P = r. Hence, the angular momentum at P is m(r θ̇ )r.
Conservation of angular momentum gives,
mv0 b = mr 2 θ̇
or
(1.36)
θ̇ = v0 b/r 2
(1.37a)
v0 b dr
dr
θ̇ = 2
dθ
r dθ
(1.37b)
Also
ṙ =
Using (1.37a) and (1.37b) in (1.35) and dividing by b2 , we find:
1 dr 2
1
2zZe2
1
+
=
+ 2
b2 mv02 b2 r r 4 dθ
r
(1.38)
22
1
Passage of Charged Particles Through Matter
It is desirable to have a change of variable, i.e.
1
r
dr
dr du
1 du
=
=− 2
dθ
du dθ
u dθ
u=
(1.39)
(1.40)
Also, it is convenient to call
R0 =
2zZe2 zZe2
=
T0
mv02
(1.41)
Here T0 is the initial kinetic energy of the incident particle. R0 is the distance of
closest approach for the head-on collision (b = 0). At the distance R0 , the particle
momentarily comes to rest (v = 0) before it makes a sharp U-turn. Using (1.39),
(1.40) and (1.41) in (1.38) and re-arranging, we get:
2
du
1
R0
= 2 − u2 − 2 u
(1.42)
dθ
b
b
Above differential equation can be solved easily if it is differentiated once with
respect to θ , and bearing in mind that b is a constant for a given encounter,
2
du R0 du
d 2 u du
= −2u
− 2
2
dθ
dθ dθ
b dθ
Cancelling the common factor du/dθ and re-arranging we get:
R0
d 2u
+u+ 2 =0
2
dθ
2b
(1.43)
This has the obvious solution
u = A cos(θ − δ) −
R0
2b2
(1.44)
where A and δ are the constants of integration. We may choose δ = 0 to make the
trajectory symmetrical about the x-axis. Call
g = 2b2 /R0
(1.45)
1
g
=r =
u
gA cos θ − 1
(1.46)
with δ = 0, we find from (1.44)
This may be compared with the equation for a conic
r=
a(ε 2 − 1)
ε cos θ − 1
(1.47)
1.3 Rutherford Scattering
23
where ε is the eccentricity and ‘a’ is the semi-major axis. We therefore, identify
g = a ε2 − 1
(1.48)
ε = gA
(1.49)
Using (1.44) and (1.45) in (1.42) and simplifying
A2 =
1
1
+ 2
2
g
b
Eliminating A between (1.49) and (1.50) and using (1.48) we can find ε
4b2 T 2
4b2
ε = 1 + 2 = 1 + 2 2 04
z Z e
R0
(1.50)
(1.51)
where we have used (1.41).
It is seen from the above formula that ε > 1 even if the charge is negative and the
eccentricity is same in both the cases. The orbit is always a hyperbola and never an
ellipse. For a repulsive Coulomb force (positively charged incident particle) the orbit
is a hyperbola with the target nucleus at the external focus F , whereas for attractive
Coulomb force (negatively charged incident particle) the orbit is a hyperbola with
the target nucleus at the inner focus F .
As r → ∞, the denominator of the right hand side of (1.47) becomes zero, and
the limiting angle α is given by:
cos α =
1
ε
or
1
(1.52)
cot α = √
ε2 − 1
Observe that α is very nearly equal to half of the angle subtended between the
asymtotes, since the angle contained between the radius vector r and the x-axis is
almost equal to α when r → ∞. The scattering angle θ0 is equal to angle BOD and
is given by θ0 = π − 2α or, α = π2 − θ20 . Hence
tan
1
R0
θ0
= cot α = √
=
2
2
ε − 1 2b
(1.53)
where we have used (1.52) and (1.51).
Formula (1.53) can be derived by a shorter method by assuming that the trajectory is a hyperbola and by considering the velocity v at the point C which is at
distance ‘a’ from the centre of the nucleus (Fig. 1.20); v being perpendicular to ‘a’.
Energy conservation gives:
1 2 1 2 zZe2
mv = mv +
2 0 2
a
24
1
Passage of Charged Particles Through Matter
which yields
v
v0
2
=1−
R0
a
(i)
Angular momentum conservation gives
mv0 b = mva
or
b
v
=
v0 a
(ii)
Using (ii) in (i), we find
b2 = a(a − R0 )
(iii)
From the properties of hyperbola, we know
a = b cot
α
2
(iv)
Eliminating ‘a’ between (iii) and (iv)
cot2 α2 − 1
R0
= 2 cot α
=
b
cot α2
cot α = tan
θ 0 R0
=
2
2b
Equation (1.53) shows that smaller the impact parameter b, larger is the scattering
angle θ0 , and vice versa. Physically a larger value of b implies a weaker force and
so a smaller deflection is to be expected.
In particular, b = ∞, implies θ0 = 0 and b = 0 implies θ0 , = π . Figure 1.21
shows three typical scattering events. They are:
(a) with a large b,
(b) with a moderate value of b, and
(c) for a very small value of b.
Eliminating g between (1.45) and (1.48)
a=
2b2
R0
=
2
R0 (ε 2 − 1)
(1.54)
where we have used (1.51).
For a particular value of b, the closest distance of approach will be F C which is
given by putting θ = 0 in (1.47)
R0
4b2
r(min) = a(ε + 1) =
1+ 1+ 2
(1.55)
2
R0
1.3 Rutherford Scattering
25
Fig. 1.21 Rutherford scattering for three different parameters b
Considering various scattering events with different b, r(min) will take on the least
value for b = 0, i.e. for the head-on collision and in this case r(min) = R0 . Thus
the significance of R0 given by (1.41) is that it represents the least distance of the
closest approach. It is also called Collision diameter. This result also follows from
very simple considerations. As the positively charged particle approaches the target
nucleus, due to the Coulomb’s repulsion, it loses kinetic energy and when it is closest to the nucleus in a head-on collision, it would lose all its kinetic energy. Putting
v = 0 in (1.33) we get
r(min) =
2zZe2
= R0
mv02
We can find the numerical value of R0 in the scattering of 5 MeV alpha particles
(typical alpha energy from the radioactive sources) from a gold foil
T0 = 5 MeV = 5 × 1.6 × 10−13 J = S × 10−13 J
e = 1.6 × 10−19 Coul
z = 2,
Z = 79
zZe2
4πε0 T0 .
1.44zZ
1.44×2×79
T0 (MeV) =
5
For S.I. units, formula (1.41) becomes: R0 =
For numerical calculations, R0 (fm) =
= 45.5 fm.
This value may be compared with the radius of gold nucleus which is 8 fm,
a value which is smaller than the minimum distance of closest approach. This then
ensures that the alpha particle of 5 MeV stays well outside the gold nucleus in any
type of encounter including the head-on collision and that the inverse square law
would be valid for all the orbits. For much greater energy, in close encounters, alpha
particles may be able to penetrate the target nucleus itself in which case the inverse
square law would no longer be valid, and other complications would be introduced
into which we shall not enter at the moment.
From (1.53) it is obvious that given the impact parameter b, the scattering angle
θ0 can be determined. But, in practice, it is impossible to know the value of b. However, we can compute the expected angular distribution from the entire range of b’s.
Consider a uniform beam of particles fired against the target material. The beam intensity I is defined as the number of particles crossing unit area normal to the beam
26
1
Passage of Charged Particles Through Matter
Fig. 1.22 Particles passing
through the ring of radii b and
b + db are scattered in the
angular interval θ0 + dθ0
and θ0
direction per second. Near the centre of force a beam particle bends around and as
it escapes from the field of force it once again describes a straight line. Because
the force is central, one can expect an azimuthal symmetry in scattering about an
axis along the beam direction. Assuming that the scattering is independent of β the
element of solid angle becomes dΩ = 2π sin θ0 dθ0 . Consider a uniform beam of
particles of same energy directed towards a force centre, with impact parameters b
and b + db. Such particles are seen to pass through a ring of radii b and b + db,
the ring being perpendicular to the beam direction and symmetrical about an axis
passing through the nucleus, and has an area of 2πbdb. Now, particles of given energy and impact parameter have a unique angle of deflection determined by formula
(1.53). Therefore, particles passing through this ring must be scattered into the solid
angle lying between θ0 and θ0 + dθ0 (Fig. 1.22). Since the number of particles must
be conserved
2πbdbI = −2π sin θ0 dθ0 I σ (θ0 )
σ (θ0 ) = −
bdb
sin θ0 dθ0
or
(1.56)
The negative sign is introduced in (1.56) due to the fact that an increase in b implies
a decrease in θ0 . Rewriting (1.53)
b=
θ0
R0
cot
2
2
(1.57)
Hence
db = −
R0
θ0
cosec2 dθ0
4
2
(1.58)
Using (1.57) and (1.58) in (1.56) and noting that sin θ0 = 2 sin θ20 · cos θ20 , we get
after simplification:
σ (θ0 ) =
R02
16 sin4
θ0
2
=
1 zZe2 2 1
16 T0
sin4 θ0
2
1.3 Rutherford Scattering
27
Fig. 1.23 Differential
cross-section (in arbitrary
units) as a function of
scattering angle θ in the LS
1 zZe2 2 1
σ (θ ) =
4 mv02 sin4 θ0
2
(Rutherford’s scattering formula)
(1.59)
This is the famous Rutherford’s scattering formula. Henceforth, the suffix 0 is
dropped off in θ0 . The expected differential cross-section as a function of scattering angle given by (1.59) is shown in Fig. 1.23. Observe that the differential
cross-section falls off rapidly with increasing angle, the scattering thus being predominantly in the forward direction.
Formula (1.59) also shows that σ (θ ) will be greater for targets and incident particles of higher atomic number and that it will be more important for low energy
particles.
For the purpose of numerical calculations (1.59) can be written in the form:
σ (θ ) = 1.295
zZ
T
2
1
4
sin θ/2
Mb/sr
(1.60)
where T is in MeV.
1.3.2 Darwin’s Formula
Rutherford’s formula which takes into account the recoil of the nucleus is due to
Darwin (see Example 1.18)
σ (θ ) =
zZe2
mv 2
2
1
[cos θ ± (1 − γ 2 sin2 θ )1/2 ]2
sin4 θ
(1 − γ 2 sin2 θ )1/2
(1.61)
where M is the mass of the target nucleus and m is the mass of the incident particle,
and γ = m/M. If γ < 1, the positive sign should be used only before the square
root. If γ > 1 the expression should be calculated for positive and negative signs
and the results are added to obtain γ (θ ). For γ = 1
zZe2
σ (θ ) =
T
2
cos θ
sin4 θ
(1.62)
28
1
Passage of Charged Particles Through Matter
1.3.3 Mott’s Formula
If the scattered and the scattering particles are identical (Indistinguishable particles),
the quantum mechanical exchange effects must be taken into account. The scattering
formula due to Mott is
z2 Z 2 e4 cos θ
1
1
+2
1
cos γ 2 Z 2 e2
2
σ (θ ) =
ln tg θ
+
v
T2
cos4 θ −1 sin2 θ cos2 θ
sin4 θ
(1.63)
where h is Planck’s constant. +2 is put infront of the square brackets if the particles
have zero spin, and −1, if their spin is 12 .
1.3.4 Cross-Section for Scattering in the Angular Interval θ and θ The cross-section σ (θ , θ ) per nucleus for scattering between angle θ and θ is
given by:
σ θ , θ =
θ θ
σ (θ )dΩ = 2π
θ θ
2πR02
sin θ σ (θ )dθ =
16
θ θ
sin θ dθ
sin4 θ/2
where (1.59) has been used. As sin θ = 2 sin θ2 cos θ2
π
σ θ , θ = R02
4
π
= R02
2
θ θ
cosec2 (θ/2) cot(θ/2)dθ
θ θ cot(θ/2)d cot(θ/2)
(1.64)
1
σ θ , θ = πR02 cot2 θ /2 − cot2 θ /2
4
In particular, σ (90◦ , 180◦ ) the cross-section for scattering for angles greater than
90° is given by setting θ = 90◦ and θ = 180◦ in (1.64)
πR02 π zZe2 2
σ 90◦ , 180◦ =
=
(1.65)
4
4
T0
1.3.5 Probability of Scattering
Consider a box of face area 1 cm2 and length λ cm, so that its volume becomes
λ cm3 . Let a beam of particles be incident on its face. By definition λ is such a
1.3 Rutherford Scattering
29
Fig. 1.24 A box of face area
1 cm2 and length λ cm
containing n atoms is exposed
to a beam of particles
length that on an average the particle suffers the given type of scattering, i.e. λ is the
mean-free-path. If there are n number of atoms per cm3 , the number of atoms inside
the box of volume λ cm3 will then the equal to λn. The cross-section arising from all
these atoms will then be equal to λnσ (θ , θ ). Imagine all the atoms inside the box
to be pushed on the rear surface of the box (Fig. 1.24). The total area corresponding
to the cross-section of all the atoms must be such as to completely fill up area of
1 cm2 since our assumption demands that on an average one scattering of the given
type will occur when the incident particle passes through λ cm
∴ λnσ θ θ = 1 or
1
nσ θ , σ =
λ
If the foil is only t cm thick, then the probability of scattering between θ and θ will be:
P = t/λ = ntσ θ, θ (1.66)
1.3.6 Rutherford Scattering in the LS and CM System
So far, we have considered the scattering of particles from massive target nuclei so
that the recoil of the latter can be neglected altogether. However, if a light target
be considered then the target nucleus would necessarily recoil due to the collision
and the analysis of the collision is rendered fairly complicated when done in the
lab system. Figure 1.25 shows for definiteness the elastic scattering of an α-particle
(m1 = 4) with a carbon nucleus (m2 = 12) originally stationary seen in the lab system. The α particle moves with velocity u1 , and makes an impact parameter b. Since
m2 is assumed to be stationary, the relative velocity of approach is also u1 .
The centre of mass (indicated by CM in the diagram) moves with constant velocity vc = m1 u1 /(m1 + m2 ), before, during and after the collision, which is always directed parallel to the incident direction of m1 . In the chosen example, vc is
one-fourth of the initial velocity of the α particle. We have seen that the analytical
relationships which connect the scattering angle θ and φ with the impact parameter
b and with the charges, masses and velocities of m1 and m2 are too complicated
to be of any general use. Observe that after the collision the initial direction of m2
is away from that of m1 . This is a simple consequence of Coulomb’s repulsion between the two nuclei. It must be pointed out that the trajectories are no longer simple
30
1
Passage of Charged Particles Through Matter
Fig. 1.25 Scattering of α
particles with a carbon
nucleus in the LS
Fig. 1.26 Scattering of α
particles with a carbon
nucleus in the CMS
hyperbolas in the lab system. In the CM system, no distinction is made between the
projectile and the target particles, see Fig. 1.26. The relative velocity of the particles
is, v(rel) = u∗1 + u∗2 = (u1 − uc ) + uc = u1 , which is identical with that in the lab
system.
There is complete symmetry in the scattering of the particles in the CM system. Both the particles approach each other with equal and opposite momentum
before the collision and recede with equal and opposite momentum after the collision. In the event of elastic scattering, the respective speeds of the particles remain
unaltered before and after the collision. Both are deflected through the same angle
measured with their respective original direction. Their centre of mass remains at
rest throughout the collision. Each of the particles describes a hyperbola. The collision diameter, impact parameter and eccentricity of the orbit are the same for both
the particles, Fig. 1.27. In our example, α particle traverses its hyperbolic path r1
about the centre of mass, while the carbon nucleus also traverses a similar path,
r2 = r1 m1 /m2 = r1 /3, on the other side of the centre of mass. The line joining the
positions of the two particles passes through the centre of mass at all times. The
angular momentum about the centre of mass evaluated in the CM system (Fig. 1.27)
1.3 Rutherford Scattering
31
Fig. 1.27 Angular
momentum about the centre
of mass in the CMS
is
J = m1 (u1 − vc )
=
r1 b
m2 vc r2 b m1 u1 r1 b
vc b
+
=
+
(m2 r2 − m1 r1 )
r1 + r2
r1 + r2
r1 + r2
r1 + r2
m1 u1 r1 b
r1 + r2
Since m1 r1 = m2 r2
r1
m2
=
r1 + r2
m1 + m2
J = m1 u1
bm2
= μu1 b
m1 + m2
where μ is the reduced mass. The angular momentum J of this system of two particles is a constant of their motion since no external torques act on the system. The
angular momentum taken about the centre of their mass has the same value both in
the lab system and CM system since these two systems differ only in regard to the
translation velocity of the centre of mass (vc in the lab system and zero in the CM
system).
1.3.7 Validity of Classical Description of Scattering
We must be able to form a wave packet which is narrower than the distance of the
closest approach, otherwise there is no way to make sure that the particle experiences a definitely predictable force from which the deflection can be calculated
classically. To obtain a rough estimate of the validity of the classical description,
we can safely assume that the distance of closest approach is of the same order
of magnitude as the impact parameter b. In order to form a wave packet that is
smaller than b, it is of course necessary that one uses a range of wavelengths of
the order of b or smaller. Thus the first requirement is that the momentum of the
incident particles be considerably larger than p = /b. Moreover, in defining the
position of this packet will make the momentum of the particle uncertain by a quantity much greater than δp = /b. This uncertainty will cause the angle of deflection
to be made uncertain by a quantity much greater than δθ = δp/p. In order that the
classical description be applicable, the above uncertainty ought to be a great deal
smaller than the deflection itself; otherwise the entire calculation of the deflection
by classical method will be meaningless. This requirement is, however, equivalent
32
1
Passage of Charged Particles Through Matter
to the requirement that the uncertainty in the momentum be much smaller than the
net momentum, p, transferred during the collision, or that
δp/
p = /b
p 1
(1.67)
Now, for elastic scattering, for small angles
p = 2p sin(θ/2) pθ
(1.68)
θ = zZe2 /T b
(1.69)
Also from (1.53)
Combining (1.68) and (1.69) and noting that p/T = 2/v, the condition is,
2zZe2 /v 1
or
2zZ/137β 1
(1.70)
For 5 MeV α and gold nucleus (Z = 79) as the target, β = 0.05, and the left hand
side of (1.67) becomes 46, a value which is much greater than unity, so that classical
description of scattering is fully valid.
1.3.8 Coulomb Scattering with a Shielded Potential Under Born’s
Approximation
It is always an abstraction to assume that the Coulomb force continues to be unmodified out to arbitrarily large distance. Thus the Coulomb force resulting from
distances of the order of a few atomic radii is screened or shielded by the atomic
electrons. The resulting shape of the potential may be approximated by the shielded
Coulomb potential of the form
V=
zZe2
exp(−r/r0 )
r
(1.71)
The exponential factor causes the force to become negligible when the factor r/r0
is much greater than unity. According to the Born approximation, the expression for
the differential cross-section is given by
σ (θ ) =
2
4π 2 m2 V (p − p0 )
h4
(1.72)
where the momentum transfer is
|p − p0 | = 2p sin(θ/2)
(1.73)
and
V (p − p0 ) = zZe2
4πzZe2
exp(−r/r0 )dr
= p−p0
exp i(p − p0 ) · r
r
| 2 |2 +
1
r02
(1.74)
1.3 Rutherford Scattering
33
Fig. 1.28 Rutherford
scattering with a shielded
potential
Letting r0 → ∞, and combining (1.72), (1.73) and (1.74), we get exactly the same
expression as (1.59), i.e. Rutherford scattering law. Thus, classical mechanics and
quantum mechanics give the same result for the Rutherford scattering.
The general appearance of the cross-section for a shielded Coulomb force as a
function of angle is shown in Fig. 1.28. The curve rises steeply with decreasing θ ,
as is characteristic of the Rutherford cross-section, until
sin
θ0 ∼ =
2
2pr0
For angles smaller than θ0 , the rise of σ (θ ) is comparatively small. Thus, θ0 , may be
regarded as a sort of minimum angle, below which Rutherford scattering ceases, as
a result of the shielding effects. With a shielded Coulomb potential, θ0 will approach
zero with increasing b much more rapidly, as soon as b goes beyond the shielding
radius. In fact, shortly beyond the shielding radius, the entire scattering effect can
be neglected. The minimum angle below which the cross-section ceases to increase
is given by setting b = r0 in Eq. (1.69) adapted for small angle approximation,
θ (min) = zZe2 /T r0 .
1.3.9 Discussion of Rutherford’s Formula
The formula fails for indistinguishable particles and also for relativistic particles.
The derivation ignores spin interaction. Formula (1.59) predicts pronounced scattering in the forward direction, i.e. small angle scattering is favoured. Spin interaction,
however, affects only the large angle scattering. Screening effect of electrons has
been ignored which tends to reduce the effective charge of the nucleus. This effect
is small for small impact parameters (large θ ) and will clearly manifest itself in
heavy atoms in which K electrons are very close to the nucleus.
Rutherford’s formula is valid only for single scattering. It is, therefore, necessary
to use thin foils; otherwise, multiple scattering will result from the superposition
of successive single scatterings. Scattering due to orbital electrons may be ignored
since maximum angle of scattering will be typically <10−4 radians.
If the mass of the incident particle cannot be neglected in comparison with the
mass of the target nucleus, then (1.59) is still valid provided the energy T now refers
to the centre of mass system, and similarly the angle.
34
1
Passage of Charged Particles Through Matter
The total cross-section σ is given by
π
σ = 2π
σ (θ ) sin θ dθ
(1.75)
0
By inserting (1.59) in (1.75), it is seen that σ → ∞. This is the direct consequence
of the long range character of Coulomb forces. But physically, at ranges (impact
parameters) comparable with the atomic radii, the force field is effectively cut-off,
leading to a finite cross-section. In other words, divergence in σ which results at
θ = 0 does not occur in practice, since infinite impact parameters are not effective
owing to the shielding of the nucleus by the orbital electrons.
1.3.10 The Scattering of α Particles and the Nuclear Theory
of Atom
The experiments of Geiger and Marsden suggested that if alphas passed through thin
foils of platinum then about 1 in 8000 suffered a deflection larger than 90°. From
the then existing Thompson’s theory of atom (protons and electrons uniformly distributed in the atom), it appeared that the electric fields are weak, much too weak to
give rise to such deflections. It was conjectured that a cumulative effect of a large
number of small angle scatters might result in a large deflection. But it was not possible to explain deflections of this magnitude. To explain this result, Rutherford postulated that positive charge is concentrated in a very small volume called the nucleus.
It was then shown that all the wide angle scatters could be explained due to the existence of the nucleus alone, and that the screening effect due to the orbital electrons
could be ignored. Rutherford used the scintillation method to count the scattered
alphas, and an excellent agreement was made with the theory. The results were so
satisfactory that Rutherford used his formula to verify the atomic numbers of some
of the elements that were already known from Mosley’s work on X-ray spectra of elements, and an extraordinary agreement was obtained. The dependence of scattering
on energy and angle left no doubt as to the validity of Coulomb law of force, the distance of closest approach being 70 to 140 fm (1 fermi = 10−13 cm). For alpha scattering against gold foils, the scattering was well represented by the Coulomb law of
force when the closest distance of approach was 32 fm. It was, therefore, concluded
that the radius of gold nucleus is not greater than this value (actual value being
about 8 fm). Thus, any departure from Rutherford’s law of scattering is an indication that the incident particle is coming into contact with the target nucleus. Careful
experiments can, therefore, give reasonably good estimates of nuclear sizes. For
the range of energies considered (alphas from radioactive substances), no departure
from Coulomb inverse square law of forces was observed, for very close distance
of collision. The alpha scattering was found to be strictly governed by the Rutherford scattering law for targets ranging from copper to uranium. In the mean time,
Blacket obtained an independent verification from Wilson chamber photographs by
1.3 Rutherford Scattering
35
determining the angular distribution at various energies. But experiments with light
elements revealed serious departure from the Rutherford scattering law at large angles. Observe that R0 (minimum distance of approach) is proportional to Z (charge
1
of the target nucleus) while the nuclear radii are roughly proportional to Z 3 for
light and medium elements. The latter, therefore, decreases much slower than the
former. Consequently, alphas could penetrate the nuclei of light elements. It was
also observed that when alphas of larger energies were employed, a more rapid departure from the Rutherford law of scattering was approached, the observed number
being greater than the predicted number. These results confirmed the general result
that when the particles approach close to the nucleus the simple Coulomb law is no
longer obeyed and that no known law concerning the electrical forces could explain
the observed scattering. Explanation in terms of magnetic interaction between the
alpha particle and the nucleus was ruled out since it was already known that alpha
particle has zero spin, and therefore does not have any magnetic moment. Further,
the validity of the classical formula of Rutherford could not be questioned since
quantum mechanics also gives precisely the same result. The only answer to the
anamolous scattering was to postulate the existence of strong short range attractive
nuclear forces. For large b the angle θ will be given by pure Coulomb forces. As b
decreases θ would increase, but not so rapidly as in the pure Coulomb field. This
has the consequence of more particles to be scattered at small angles and fewer at
large angles. As b decreases further, the attractive nuclear forces may more than
compensate for the repulsive Coulomb forces. The angle of scattering may therefore decrease rather than increase, and in this case there will be a maximum angle
of scattering. For very small values of b, the particles may get scattered at random
and in some cases may get captured.
Example 1.7 If σg is the geometrical cross-section for uncharged particles for hitting a nucleus, show that for positively charged particles, the cross-section will decrease by the factor (1 − RR0 ), where R0 = zZe2 /4πε0 K0 R = radius of the nucleus,
σg = πR 2 and K0 is the kinetic energy.
Solution When the charged particle just grazes the nucleus
1
4b2
r(min) = R = R0 1 + 1 + 2
2
R0
(1.55)
whence we obtain b2 = R 2 − RR0 .
Denoting the cross-section by σ = πb2 , it follows that σ/σg = πb2 /πR 2 = 1 −
R0 /R.
Example 1.8 Alphas of 8.3 MeV bombard an aluminum foil. The scattered alphas
are observed at an angle of 60°. Calculate the minimum distance of approach in this
case.
36
1
Solution
Passage of Charged Particles Through Matter
T0 = 8.3 MeV; z = 2, Z = 13
θ
60
R0
1
= tan = tan
= tan 30◦ = √
2b
2
2
3
R0
4b2
r(min) =
1+ 1+ 2
2
R0
(1)
(2)
Using (1) in (2)
r(min) = 3R0 /2
R0 =
1.44zZ 1.44 × 2 × 13
= 4.51 fm
=
T0
8.3
r(min) = 1.5 × 4.51 = 6.76 fm
Example 1.9 Find the probability of scattering of alpha particles of energy 5 MeV
through an angle greater than 90° in their passage through a foil of gold of thickness 4 × 10−5 cm. Given, Avogadro’s number N0 = 6 × 1023 ; A = 196; Z = 79;
electronic charge e = 1.6 × 1019 C; 1/4πε0 = 9 × 109 N m2 /C2 , density of gold =
19.3 g/cm3 .
Solution
N = Number of atoms/cm3 = (number of atoms/g) g/cm3
6 × 1023 × 19.3
= 5.91 × 1022
196
π 1.44zZ 2 π 1.44 × 2 × 79 2
◦
◦
σ 90 , 180 =
=
4
T0
4
5
=
= 1625 fm2 = 16.25 × 10−24 cm2
If λ is the mean-free-path, λ1 = N σ = 5.91 × 1022 × 16.25 × 10−24 = 0.961 cm−1 .
Required probability p = t/λ = 4 × 10−5 × 0.961 = 3.84 × 10−5 . In other words,
one alpha in 1/(3.84 × 10−5 ) or 26000, gets scattered at an angle greater than 90°.
Example 1.10 If the radius of gold nucleus (Z = 79), is 8 × 10−15 m, what is the
minimum energy that the a particle should have to just reach it? Give your answer
in MeV.
Solution Set
R = R0 = 8 fm
T =
1.44zZ 1.44 × 2 × 79
= 28.44 MeV
=
R0
8
1.3 Rutherford Scattering
37
Example 1.11 The following counting rates (in arbitrary units) were obtained when
a particles were scattered through 180° from a thin gold (Z = 79) target. Deduce a
value for the radius of a gold nucleus from these results.
Energy of α particle
(MeV)
8
12
18
22
26
27
30
34
Counting rate
30300
13400
6000
4000
2800
33
4
0.4
Solution Since at a given angle, the counting rate is inversely proportional to the
square of energy, we can calculate the expected counting rate for various energies,
assuming that at 8 MeV (lowest energy) the counting rate N8 is in agreement with
the expected value. In the table below are displayed the calculated counting rates
with the aid of the formula, N = N8 (8/T )2 , where T is in MeV.
T in MeV
8
12
18
22
26
27
30
34
N (cal)
N (obs)
30300
30300
13100
13400
5990
6000
4010
4000
2867
2800
2667
33
2157
4
1679
0.4
Comparison between the calculated and observed counting rates indicates that departure from Rutherford scattering begins at 26 MeV. Since scattering angle is
θ = 180◦ , we are concerned with head-on collisions. Hence
R0 =
1.44zZ 1.44 × 2 × 79
= 8.75 fm
=
T0
26
Hence, the radius of the gold nucleus is 8.75 fm.
Example 1.12 (a) If a gold foil is bombarded by 5.4 MeV a particles, determine the
distance of closest approach (b) what is the deflection of the alpha particle when the
impact parameter is equal to this distance?
Solution
(a) Distance of closest approach
R0 =
1.44zZ 1.44 × 2 × 79
= 42.1 fm
=
T0
5.4
(b)
tan
θ
R0
R0
1
=
=
=
2
2b
2R0 2
θ = 53◦ 8
38
1
Passage of Charged Particles Through Matter
Example 1.13 To what minimum distance will an alpha particle of energy 0.4 MeV
approach a stationary Li7 nucleus in the case of a head-on collision? Take the nuclear
recoil into account.
Solution We work out in the CM system. Equating the potential energy at the
closest distance of approach R0 to the initial K.E.
T0 m2
1.44zZ
1
1 m1 m2 v 2
=
= μv 2 =
or
R0
2
2 (m1 + m2 ) m1 + m2
m1
1.44 × 2 × 3
4
1.44zZ
1+
=
R0 =
1+
= 33.9 fm
T0
m2
0.4
7
Example 1.14 A narrow beam of alpha particles with kinetic energy T = 600 keV
falls normally on a golden foil incorporating n = 1.1 × 1019 nuclei/cm2 . Find the
fraction of alpha particles scattered through the angles θ < θ0 = 20◦ .
Solution Given nt = 1.1 × 1019 nuclei/cm2
π 1.44zZ 2 2 θ
π 2 2θ
· nt = 1 −
· nt
N/N = 1 − R0 cot
cot
4
2
4
T0
2
π 1.44 × 2 × 79 2 2 20
=1−
× 1.1 × 1019 × 10−26 = 0.6
cot
4
0.6
2
The factor 10−26 has been introduced to convert fm2 into cm2 .
Example 1.15 A narrow beam of protons with kinetic energy T = 1.4 MeV falls
normally on a brass foil whose mass thickness ρt = 1.5 mg/cm2 . The weight ratio
of copper and zinc in the foil is equal to 7 : 3, respectively. Find the fraction of the
protons scattered through the angles exceeding θ = 30◦ .
Solution
2
0.3Z22
π 10−26
N
θ
2 0.7Z1
ρtN0 cot2
=
× 1.44
+
N
4 T2
M1
M2
2
where Z1 and Z2 are the atomic numbers of copper and zinc, M1 and M2 are their
molar masses, N0 is the Avagadro’s number. The factor 10−26 is introduced to express fm2 as cm2
(1.44)2 0.7 × 292 0.3 × 302
N
=
+
× 1.5 × 10−3 × 6 × 1023
N
63.55
65.38
(1.4)2
× (3.732)2 × 10−26
= 1.4 × 10−3
1.3 Rutherford Scattering
39
Example 1.16 Find the effective cross-section of a uranium nucleus corresponding
to the scattering of alpha particles with kinetic energy T = 1.5 MeV through the
angles exceeding θ = 60◦ .
Solution
π
σ (θ, π) =
4
1.44zZ
T
2
π
θ
=
cot
2
4
2
1.44 × 2 × 92
1.5
2
cot2 30◦
= 73480 fm2 = 735 b
Example 1.17 The effective cross-section of a gold nucleus corresponding to the
scattering of monoergic alpha particles within angular interval from 90° to 180°
is equal to σ = 0.5 kb. Find (a) the energy of alpha particles (b) the differential
cross-section of scattering σ (θ ) (kb/sr) corresponding to the angle θ = 60◦ .
Solution
(a)
◦
π 1.44zZ 2
◦
σ 90 , 180 =
4
T
√
√
π × 1.44zZ
π × 1.44 × 2 × 79
=
T=
= 0.9 MeV
√
√
2 × σ (90 , 180 )
2 × 5 × 104
(b)
2
dσ
zZ
1
= 1.295
dΩ
T
sin4
θ
2
2 × 79
= 1.295
0.9
2
1
sin4
6
2
= 0.638 × 106 mb/sr = 0.64 kb/sr
Example 1.18 Derive Darwin’s formula for scattering (modified Rutherford’s formula which takes into account the recoil of the nucleus).
Solution
3
σ (θ ) =
(1 + γ 2 + 2γ cos θ ∗ ) 2 ∗ σ θ
|1 + γ cos θ ∗ |
(1)
Now, Rutherford’s formula for CMS is
1 zZe2 2 1
σ θ∗ =
∗
4 μv 2
sin4 θ2
(2)
40
1
Passage of Charged Particles Through Matter
Also
sin4
sin4 θ ∗
θ∗ 1
=
2
4 (1 + cos θ ∗ )2
and
m
mM
=
m+M
1+γ
μ=
(3)
(4)
where M and m are the target and incident particle masses respectively and γ =
m/M
tan θ =
sin θ ∗
γ + cos θ ∗
(5)
Squaring (5) and expressing it as a quadratic equation and solving it we get
∗
cos θ = γ sin θ ± cos θ 1 − γ 2 sin2 θ
2
(6)
combining (1), (2), (3), (4) and (6), and after some algebraic manipulations we get
σ (θ ) =
zZe2 2
mv 2
1 − γ 2 sin2 θ ]2
[cos
θ
±
1
sin4 θ
1 − γ 2 sin2 θ
(7)
This is Darwin’s formula. For m M, γ → 0 and (7) reduces to the usual Rutherford formula.
1.4 Multiple Scattering
1.4.1 Mean Scattering Angle
A charged particle in passing through a thick medium is scattered through an angle θ . The observed scattering may be the result of cumulative effect of a number of
small deflections produced by different atomic nuclei in the matter traversed, or it
may be a single deflection through an angle θ produced by a single nucleus. The first
type of scattering is called multiple or plural, according to the number of contributing collisions is large or small. The second type is referred to as single scattering.
Which process is mainly operative depends on the nature and velocity of the scattered particle, the matter traversed and the scattering angle. In the simple treatment,
it is assumed that Θ is distributed about Θ = 0, according to the Gaussian law, i.e.
the probability for scattering in the angular interval Θ and Θ + dΘ is
P (Θ)dΘ = const · exp −KΘ 2 dΘ
(1.76)
1.4 Multiple Scattering
41
Fig. 1.29 Multiple scattering
Θi resulting from the
superposition of single
scatterings, θ1 , θ2 , . . . , θi
where K is a constant. The single scattering is governed by the Rutherford law of
scattering, which for small angles has the form
dθ
θ3
P (θ )dθ = const ·
(1.77)
Let θi be the deflection in the ith collision. Let there be q collisions in a traversal of
t cm. Since small angles are vectors (Fig. 1.29)
q
Θq =
θi
i=1
Take the dot product
Θq2
=
q
θi2
+
q
θi θj
i=j
i=1
In averaging over many traversals, θj is positive as many times as it is negative; and
the second summation drops off. Thus
Θq2
=
q
θ12
i=1
Since statistically, the individual events do not differ, θi2 = θ 2 we can therefore write
Θ2 = q θ2
(1.78)
Now, Rutherford scattering for small angles can be written as
σ (θ ) = 4
zZe2
pv
2
1
θ4
(1.79)
The probability for the particle to be scattered into solid angle dΩ = 2π sin θ dθ , is
given by σ (θ )2πθdθ for small θ . Let there be N atoms/cm3
bmax 4z2 Z 2 e4
θmax 2
2
2
θ f (b)db
bmin p 2 v2 b2 2πNtdb
θmin θ f (θ )dθ
= =
θ = θ
max
f (b)db
2πbdbNt
f
(θ
)dθ
θmin
42
since tan θ2 =
1
zZe2
Mbv 2
Passage of Charged Particles Through Matter
and for small θ
θ2 =
4z2 Z 2 e4
p 2 v 2 b2
(1.80)
The denominator is nothing but q. We, therefore, find after substituting (1.79)
in (1.80)
2
zZe2 2 θ (max)
Θ = q θ 2 = 8πNt
(1.81)
ln
pv
θ (min)
We have assumed that the particle is sufficiently energetic so that the velocity does
not change over the considered traversal. We also ignore the scattering off the electrons, since it is unimportant. Observe that scattering off the nuclei is proportional
to Z 2 , whilst for electrons, it is proportional to Z. In hydrogen, the scattering off
electrons, however, will be important. We shall now consider the limits θ (max) and
θ (min). Nuclear scattering at large distance is reduced by the electrostatic shielding
of the nucleus by its electrons. The electrostatic shielding reduces the scattering of
distant particles but it does not reduce the energy loss. Thus, a primary particle travelling at a distance such that the fields of electrons and nuclei compensate each other
almost completely is still capable of transferring energy to the atom. The physical
reason why screening reduces scattering much more than the energy transfer is as
follows. A fast particle passing near an atom transfers a certain amount of momentum to each electron and also to the nucleus. The angle of scattering is, however,
determined by the transverse component of the total recoil. Due to the opposite sign
of charge, the electrons recoil in the opposite direction to the nucleus and if the fast
particle passes at a sufficient distance, the transverse components of the recoiling
electrons cancel the transverse components of the nuclear recoil and thus no scattering results. The total energy transfer is equal to
Z
pi2
P2
+ nuc
2me 2Mnuc
(1.82)
i=1
The energy transfer to any of the electrons or to the nucleus is positive and is not
affected by the presence of other particles except for the small effects of the binding
forces. In other words, a particle passing near an atom suffers Z + 1 collisions with
the constituents of the atom and loses energy to everyone of them. The Z + 1 angles
of scattering, however, tend to compensate each other and therefore the scattering
is reduced strongly by shielding. Now by (1.81), the root mean square multiple
scattering angle is given by
Θ2 =
√
tze
8πZ 2 z2 e4 N t b(max)
=
K
ln
b(min)
pv
p2 v 2
(1.83)
1.4 Multiple Scattering
43
where
K=
8πNZ 2 e2 ln
b(max)
b(min)
(1.84)
is called the scattering constant.
1.4.2 Choice of b(max) and b(min)
We may choose
b(max) =
ao
Z 1/3
(1.85)
where a0 is the Bohr radius. This is justified by Fermi-Thomas model of the atom,
since the right hand side of (1.85) represents the radius of the atom.
The limit on b(min) is dictated by the finite size of the nucleus. Thus, b(min) is
greater than 1.3 × 10−13 A1/3 cm. An alternative criterion would be to avoid counting deflections with Θ > 1 radian. A rough criterion is provided by restricting the individual single scatterings to θ < 1. Now for small scattering angles, formula (1.53)
reduces to
θ=
2zZe2
mv 2 b
(1.86)
Putting θ = 1 radian and p = mv, we obtain the rough criterion
b(min) =
2zZe2
pv
(1.87)
Fortunately, the results are insensitive to the choice of b(max) and b(min)
b(max)
Atomic dimension
10−8 cm
= −12
= 104 ,
b(min)
Nuclear dimension 10
cm
and so ln 104 10
Thus, Θ2 is very insensitive to the logarithmic term in (1.83) which is of the
order of 10. The main dependence comes from the factors outside the logarithmic
term. Further, in view of (1.87) the scattering
constant K is a slow varying function
2
of the particle velocity. Observe that Θ is directly proportional to the charge of
the scattering nuclei, and the charge of the incident particles, inversely proportional
to the energy of the incident particles and directly proportional to the square root of
the thickness of the absorber.
44
1
Passage of Charged Particles Through Matter
1.4.3 Mean Square Projected Angle and the Mean Square
Displacement
Consider a charged particle traversing an absorber of thickness t. Assume that the
collisions take place at depths X1 , X2 , . . . , resulting in deflections θ1 , θ2 , . . . . The
azimuth of the deflection will change after each collision, the subsequent azimuths
being φ1 , φ2 , . . . , the projected angle of deflection is given by
ΘP =
q
θi cos φi
i=1
We have to average over the parameters of the single collisions. Since the azimuths
can be taken as independent, we have
1
cos φi cos φj = δij
2
(1.88)
where δij is the Kronecker delta. Hence
1 1 ΘP2 = q θ 2 = Θ 2
2
2
(1.89)
Observe that the most probable value of Θ or ΘP is zero. However, Θ and Θ 2 are necessarily positive, whereas ΘP is zero.
Similarly the mean square projected displacement is equal to half the mean
square unprojected displacement
2 1 2
y = r
2
Now, y =
q
i=1 Xi θi
cos φi
∴
y2 =
xi xj θi θj cos φi cos φj
i=j
Since x and φ are independent, using relation (1.88)
2 1 2 2 xi θ
y =
2
i
Now
2 2 1
xi = X =
t
2 t 2
y = q;
6
t
0
x 2 dx =
t2
3
2 t 2 2
θ =
Θ
6
(1.90)
1.4 Multiple Scattering
45
Fig. 1.30 Angular distribution of electrons scattered from AU at 15.7 MeV. Solid lines indicate the
distribution expected from the Moliere theory for small-and large-angle multiple scattering, with
an extrapolation in the transition region: dashed lines, the distributions according to the Gaussian
and single scattering theories. The ordinate scale gives the logarithm of the fraction of the beam
scattered within 9.696 × 10−3 sr (Birkhoff)
Also
2 t 2 2
Θ
(1.91)
r =
3
Expressions (1.90) and (1.91) are of great interest in the cosmic ray shower theory
and experiments. Figure 1.30 shows the contribution from multiple scattering (Gaussian) at small angles and single scattering (Rutherford) at large angles. Kamal, Rao
and Rao (Fig. 1.31) have compared the experimental distributions of D the average
of ‘Second differences’ of 17.2 GeV/c beam tracks in photographic emulsions for
cell lengths t = 4, 6, 8, 10, 20, and 30 mm and compared with Moliere’s theory.
The quantity D is related to the projected angular deflection and is obtained from
the y-coordinates of the track; Di = yi+1 − 2yi + yi−1 , where yi is the ith coordinate of the track at constant x-intervals called cell length t. In order to avoid very
large scattering angles, a 4D cut-off is usually employed, a procedure in which all
deflections larger than four times the mean second difference are eliminated. This is
also indicated in the figure. Moliere’s probability function (Gaussian function plus
the single scattering tail) for the second differences was computed from the work of
Scott. A good agreement was found between theory and observations.
In conclusion, we may point out that Rutherford used extremely thin foils for
his classical experiments on alpha scattering in order to avoid the contribution of
multiple scattering.
From (1.83) it is clear that the determination of root-mean-square angle permits
one to estimate the energy of the particle.
Protons and electrons of the same energy will have the same root-mean-square
angle of scattering, but their ionization would be different since their velocities
would be different. Thus joint measurements of multiple scattering and ionization
46
1
Passage of Charged Particles Through Matter
Fig. 1.31 Multiple scattering
distribution for various cell
lengths, (a) 4 mm; (b) 6 mm;
(c) 8 mm; (d) 10 mm;
(e) 20 mm; (f) 30 mm
Moliere’s Gaussian function
plus single scattering tail [2]
permit us to estimate the mass of the particle and identify it. This method is particularly suitable for particles which are not too energetic and at the same time are not
brought to rest within the stack of emulsions.
The existence of multiple scattering can create problems in the curvature measurements of tracks in a cloud chamber. In certain cases the multiple scattering may
be so severe that spurious curvatures are observed even in the absence of magnetic
fields. In the case of bubble chambers, curvature measurements under magentic
fields are rendered difficult when a heavy liquid like xenon is used. It is also implied that curvature measurement in photographic emulsions under pulsed magnetic
fields are limited owing to severe multiple scattering by the heavy nuclei of silver
and bromine.
The phenomenon of multiple scattering leads to an interesting observation in
cosmic ray showers. Owing to multiple scattering in air, the electrons in the shower
undergo lateral displacement from the original path through several meters as they
traverse down the atmospheric depth (see expressions (1.90) and (1.91)).
1.5 Theory of Ionization
1.5.1 Bohr’s Formula
Charged particles in their passage through a medium lose their energy mostly
through excitation of atoms and ionization (collision) processes. The collision process is only one of several mechanisms by which charged particles may lose energy.
1.5 Theory of Ionization
47
In the case of electrons it constitutes the most important source of energy loss only
for relatively small energies. At energies of the order of 10 to 100 MeV, radiation
losses overtake the collision losses, depending on the Z of the absorber. For muons,
collision losses remain dominant up to energies of the order of 100 or 1000 MeV.
For protons, radiation losses are never significant, but the occurrence of nuclear collisions overshadows the collision losses at energies of the order of 1000 MeV or
greater. Energy loss by the emission of Cerenkov radiation is negligible except at
very high energies. Thus, in general, collision losses represent the most important
source of energy loss only for energies smaller than a certain value that depends on
the nature of the particles.
Most of the electrons ejected in ionization processes have energies very small
compared with the energy of the primary particle. Nevertheless, they are able to
produce several ion pairs before coming to rest. The total specific ionization consists of two parts (i) primary specific ionization which is defined as the average
number of ion pairs produced per g cm−2 (ii) secondary specific ionization which
refers to the average number of ion pairs per g cm−2 by all the secondary electrons and tertiary electrons and radiation. The total ionization implies the sum of (i)
and (ii). The total ionization is roughly three times the primary ionization. When
the primary particle is absorbed its energy is dissipated in exciting the atoms and
in producing secondary electrons partly by collisions and partly by radiation. The
secondary electrons radiation will excite more atoms and produce tertiary electrons
and photons and so on. It is clear that an electron will continue to lose its energy in
elastic collisions as long as its energy is in excess of the lowest excitation potential
of the atoms and the photons will be absorbed as long as their energy is greater than
the threshold energy for minimum ionization potential. In the event an atom gets
into an excited state by an inelastic collision with an electron or by the absorption
of a photon it immediately loses the excitation energy by the emission of photons
or Auger electrons. The fraction of the initial energy that is used in producing ionization is not appreciably affected by the nature of the primary particle nor by its
energy as most of the ionization and excitation processes are caused by electrons of
low energy. The classical ionization formula is originally due to N. Bohr. The basic
assumptions made in the derivation are (i) electrons are free (ii) the incident particle
remains undeviated throughout its motion.
The velocity acquired by the electron in an elastic collision is given by (1.27),
v2 = 2vc cos φ =
2vm1 cos 12 φ ∗
m1 + m2
where v = u1 . Since m2 m1 , v2 2v cos 12 φ ∗ .
The energy imparted to the electron is
1
1
T = mv22 = 2mv 2 cos2 φ ∗
2
2
(1.92)
48
1
Passage of Charged Particles Through Matter
Fig. 1.32 Atoms lying
within impact parameters b
and b + db
Since
φ∗ = π − θ ∗,
1
1
cos2 φ ∗ =
2
1 + cot2 12 θ ∗
But scattering angle in the CMS is related to the impact parameter by cot 12 θ ∗ =
2b/R0 . Hence,
T=
2z2 e4
mv 2 [R02 /4 + b2 ]
(1.93)
where we have used the fact that R0 = 2zZe2 /μv 2 and the Z = −1 for target electron, and μ = m, since the incident particle is considered much more massive than
the electron. Let there be n electron/cm3 of the medium consider a differential element of length dx along the path of the incident particle. The number of electrons
situated between the impact parameters b and b + bd over a length dx is given by
2πbdbndx (Fig. 1.32). The energy imparted to electrons for this range of b’s is
given by multiplying this number of electrons by T give by (1.93); but this is equal
to the energy lost by the primary particle by traversing the element of length dx. We
can, therefore, write:
b(max)
4πnz2 e4 bdb
4πnz2 e4
2b(max)
−dE/dx =
(1.94)
=
ln
R0
mv 2
mv 2 [b2 + R02 /4]
0
The underlying assumption that electrons are free is only approximately correct.
Actually they are bound to the atoms and can be considered free only if the collision time is short compared with the period of revolution. On the other hand, if the
collision time is long compared with the period of revolution, the electrons do not
absorb any energy at all. Let b(max) represent the impact parameter for which the
collision time τ = 1/ν, where ν is the orbital frequency of the electron
Impulse = F dt = momentum acquired by the electron
(1.95)
or
ze2 τ
= 2mT (max) =
2
b (max)
4z2 e4 /v 2 b2 (max)
(1.96)
Hence
b(max) = v/2ν
(1.97)
1.5 Theory of Ionization
−dE/dx =
49
4πnz2 e4
mv 3
(Bohr’s formula)
ln
mv 2
2ν/ze2
(1.98)
The negative sign implies that as x increases, E decreases. The quantity −dE/dx
is called the linear stopping power and is defined as amount of energy lost per unit
length in the absorber. When x is measured in g/cm2 , then this quantity is called the
mass stopping power.
Bohr’s classical formula is valid provided the particle velocity is larger than the
orbital electron velocity. The value of b(max) given in Bohr’s classical formula corresponds to such low energy transfers that they are far less than the ionization potential and are therefore incompatible with the acceptable theory of atomic structure.
For this reason, the classical theory predicts too great energy loss by high velocity
particles.
A quantum mechanical formula which is more exact is due to H.A. Bethe:
4πz2 e4 n
2mv 2
2
2
−dE/dx =
ln
− ln 1 − β − β
I
mv 2
(1.99)
where B = vc , I = mean ionization potential of the atoms of the medium; I = KZ,
K = 13.5 volts. The derivation assumes that the particle is a point charge, and that
the spin and magnetic moment are disregarded. Observe that the quantity −dE/dx
which represents ionization loss per unit length, is a function of velocity of the
particle and its charge but is independent of its mass. Bohr’s formula (1.98) is
not applicable for incident electrons for two reasons: (a) the derivation assumes
that the incident particle is undeflected during the collision which is not correct
for an electron; (b) for identical particles exchange phenomenon must be considered.
The last two terms in the bracket of (1.99) almost cancel out at low velocities
(small β). Since the logarithmic term is quite a slow varying function of velocity,
the main variation of −dE/dx comes from the factor 1/v 2 . At very low velocities,
the energy loss must go down because of the capture of the electrons by the incident
particle. This is not considered in the quantum mechanical formula which can be
relied on up to 5 MeV α’s or 1.3 MeV protons. As the velocity of the incident
particle decreases to very low values, various complicated effects enter the energy
loss mechanism. When the incident velocity becomes comparable with the K-shell
electron velocity, energy transfer to the K-shell electrons becomes difficult.
√ The
electrons effective for energy loss are those with velocities smaller than v = I /2m.
At low energies, the charge transfer process becomes more important than ionization
process. The atom or the ion formed by capturing an electron may lose the electron.
When the particle velocity is significantly greater than Bohr’s orbital velocity for the
K-shell electron, the electron loss dominates over electron capture. This corresponds
to 25 keV proton energy or 400 keV α energy.
At higher velocities, the terms ln v 2 and ln(1 − β 2 ) in the square brackets of
(1.99) become important. The ionization vs velocity curve (Fig. 1.33) passes through
a broad minimum as β → 1. The origin of the rise of ionization is due to the Lorentz
50
1
Passage of Charged Particles Through Matter
Fig. 1.33 The curve BCD gives the 1/v 2 dependence. At relativistic energies v changes little
and CD is asymptotic to v = c. At relativistic energies, the log term in v 2 /(1 − β 2 ) changes, and
increases as v → c, giving the rise at the curve from C to E. At very low energies (region AB)
Eq. (1.99) breaks down because the particle has velocity comparable to that of the orbital electrons
in the absorber, and the efficiency of energy exchange is much lower. The particle itself captures
electrons and spends part of its time reduced change
contraction of Coulomb field of the incident particle which makes possible the energy transfer to the electrons at greater distance from the particle path. At exceedingly high velocities, however, the ‘density effect’ limits the energy transfer. This is
also called the polarizability effect. In the derivation of (1.99), the atoms have been
considered isolated. This is justifiable so long the medium is a gas. In a condensed
medium, the atoms may still be considered as isolated in close collisions, but when
the impact parameter is larger than the atomic distances, the screening of the electric field due to the simultaneous movement of the electrons of the neighbouring
atoms becomes important and this has the consequence of lessening the magnitude
of −dE/dx. At ultra-relativistic velocities, the curve, therefore, gets saturated to a
plateau value called the Fermi plateau (in Fig. 1.33). It may be pointed out that in
a cloud chamber the plateau may be higher by 50 percent compared to the trough
whilst in photographic emulsions it may be higher only by about 10 percent. This is
because the density effect in the former is much less than in the latter.
1.5.2 Range-Energy-Relation
When other types of energy losses are negligible compared with the collision loss,
fluctuation in the energy loss are small and in a given material all particles of a given
energy have almost the same fixed range. The range is defined as the total distance
traversed by the particle along its track till its velocity becomes zero. In principle it
should be possible to integrate (1.99) and obtain a relation between the range and
the energy of the particle. There are two difficulties with this procedure, first the
integration is cumbersome, second at very low velocities the phenomenon of electron capture and the uncertainties in the ionization potential render the calculations
exceedingly doubtful. In practice, one uses an empirical relation of the form:
E = Kz2n M l−n R n
(1.100)
1.5 Theory of Ionization
51
where E is the kinetic energy of the particle corresponding to the range R. M is
the mass of the particle in terms of proton mass. K and n are empirical constants
that depend on the nature of the absorber. The form of (1.100) ensures that the
quantity dE/dR = z2 f (v) and = f (M), as desired. Since −dE/dx = z2 f (v),
dx = −dE/z2 f (v) = Mf (v)dv/z2 , where f (v) and f (v) are some functions of
velocity of the particle. Therefore
R
R=
0
M
dx = 2
z
v
f (v)dv =
0
M f (v)
z2
(1.101)
where f (v) is still another function of velocity.
Consider two particles of masses M1 and M2 having charges z1 and z2 . Let their
initial velocities be the same, and their ranges R1 , and R2 , respectively. It follows
from (1.101) that the expected ratio of their ranges would be
R1 M1 z22
=
R2 M2 z12
(1.102)
In particular, if the ranges of two tracks of singly charged particles from the point
of equal ionization (equal velocity) are known then
R1 M1
=
R2 M 2
(1.103)
This technique was employed for the mass determination of π meson in the historical experiment of Powell, Occhialini and Lattes, using photographic emulsions.
Comparison was made with proton tracks having the same initial ionization.
Example 1.19 The range of a low energy proton is 1500 µm in nuclear emulsions.
A second particle whose initial ionization is same as the initial ionization of proton
has a range of 228 µm. What is the mass of this particle? (The rate at which a
singly ionized particle loses energy E by ionization along its range is given by
dE/dR = K/(βc)2 MeV µm−1 where βc is the velocity of the particle, and K is a
constant depending only on emulsions; the mass of proton is 1837 mass of electron.)
Solution Using (1.103)
M2 =
R2
228 × 1837
= 279me
M1 =
R1
1500
The particle is identified as π meson (pion).
Example 1.20 α particles and deuterons are accelerated in a cyclotron under identical conditions. The extracted beam of particles is passed through an absorber. Show
that the expected range of deuterons is twice that of α particles.
52
1
Passage of Charged Particles Through Matter
Solution The condition for a circular orbit in a magnetic field (induction B) is
Bzev = mv 2 /r
Since B and r are same for both d and α
vd =
Bzer
md
and vα =
B(2e)r
mα
Since mα = 2md , it follows that vd = vα .
From (1.102)
Rd
md 22
=
=2
Rα
mα 12
Example 1.21 α-particles have an initial energy of 8.5 MeV and a range in standard
air of 8.3 cm. Find their energy loss per cm in standard air at a point 4 cm distant
from a thin source.
Solution The range-energy-relation is
E = Kz2n M 1−n R n
(1)
nE
dE
= nKz2n M 1−n R n−1 =
dR
R
(2)
Let E1 = 8.5 MeV and R1 = 8.3 cm. On moving away 4 cm from the source R2 =
8.3–4.0 = 4.3 cm. Let the corresponding energy be E2
dE2 /dR = nE2 /R2
(3)
dE1 /dR = nE1 /R1
(4)
dE2 /dR E2 R1
=
dE1 /dR E1 R2
(5)
dE2 /dR 1/v22 v12 E1
=
=
=
dE1 /dR 1/v12 v22 E2
(6)
Therefore
Also
From (5) and (6)
E1
=
E2
R1
=
R2
Using (1)
E1
=
E2
R1
R2
8.3
4.3
(7)
n
(8)
1.5 Theory of Ionization
53
Comparing (7) and (8), n = 12 . From (8) or (7)
1
2
E2 = E1 (R2 /R1 ) = 8.5
4.3
= 6.12 MeV
8.3
dE2
nE2 0.5 × 6.12
=
=
= 0.71 MeV/cm
dR
R2
4.3
1.5.2.1 Range in Air—Geiger’s Rule
If we ignore the logarithmic term in the formula for −dE/dx, then dE/dx ∝ 1/v 2
or R ∝ v 4 for the low energy region. A better approximation is provided by the
formula
R = const · v 3
(Geiger’s rule)
This formula is valid for 4–10 MeV α particles. At higher energy the exponent
changes. A Formula which gives the range of α’s in air at 15 °C and atmospheric
pressure is
R = 0.32 (MeV)3/2 cm (alphas in air)
This formula is correct to about 10 per cent in the low energy region but breaks down
for relativistic velocities. Figure 1.34 shows the range energy curves for protons and
Fig. 1.35 for alpha particles in air at 15 °C and 760 mm pressure.
1.5.2.2 The Bragg-Kleeman Rule
This rule permits one to convert the range R1 , in medium 1 of known density ρ1
and atomic weight A1 to range R2 in medium 2 of known density ρ2 and atomic
weight A2
√
R2 ρ1 A2
(Bragg-Kleeman rule)
(1.104)
=
√
R1 ρ2 A1
√
This rule is correct to within 15 per cent. As an example, for air A1 = 3.81
3
−4
−4
and
√ ρ1 = 1.226 × 10 g/cm at 15 °C, 76 cm of Hg. Then R2 = 3.2 × 10 ×
A2 R(air)/ρ2 . For aluminum A2 = 27 and ρ2 = 2.7, so that in aluminum the range
of α-particles and protons (1–10 MeV) is about 1/1600 of the range in air.
Example 1.22 Compare the stopping power of a 3 MeV proton and a 6 MeV
deuteron in the same medium.
Solution
vp =
2E
=
m
2×3 √
= 6,
1
and vd =
2×6 √
= 6
2
54
1
Passage of Charged Particles Through Matter
Fig. 1.34 Range-energy
relation for protons in air at
15 °C, 760 mm pressure up to
11.8 MeV
Since the velocities are same and also both proton and deuteron are singly charged
particles, their stopping powers are the same.
Example 1.23 Show that the specific ionization of a 320 MeV α particle is approximately equal to that of a 20 MeV proton.
Solution
−
z2
dE
∝ 2
dx
v
or
∝
Mz2
E
1.5 Theory of Ionization
55
Fig. 1.35 Range-energy
relation for alpha-particles in
air at 15 °C, 760 mm pressure
up to 15 MeV
for ∝’s,
−
1
dE 4 × 22
∝
=
dx
320
20
for p’s,
−
1
dE 1 × 12
∝
=
dx
20
20
Thus the specific ionization is same.
Example 1.24 If the range is multiplied by density, equivalent thickness in g/cm2 is
obtained. Calculate the thickness of aluminum that is equivalent in stopping power
of 1 cm of air. Given the relative stopping power for aluminum S = 1700 and its
density = 2.7 g/cm3 .
56
1
Passage of Charged Particles Through Matter
Solution
1
R(air)
=
cm
S
1700
2.7
R(Al) =
g/cm2 = 1.59 × 10−3 g/cm2
1700
R(Al) =
Example 1.25 Calculate the minimum energy an α particle can have and still be
counted with a GM counter if the counter window is made of stainless steel (A ≈ 56)
with 2 mg/cm2 thickness.
Solution For steel
Rs (cm) = Rs g/cm2 /ρs = 2 × 10−3 /ρs
Equivalent range for air
√
√
Rs ρs Aa
2 × 10−3 × 14.5
=
Ra =
√
√ = 0.83 cm
ρa As
1.226 × 10−3 × 56
0.83 2/3
R 2/3
=
= 1.89 MeV
E=
0.32
0.32
α’s of energy greater than 1.89 MeV will be counted.
Example 1.26 Calculate the range of 4 MeV α particles in air of 760 mm of Hg
pressure and 15 °C temperature.
Solution Use the formula
3
3
R = 0.32(E) 2 cm = 0.32(4) 2 = 2.56 cm
Example 1.27 Calculate the range in aluminum of a 5 MeV a particle if the relative
stopping power of Aluminum is 1700.
Solution Relative stopping power S = R(air)/R(Al). But,
3
R(air) = 0.32(5) 2 = 3.578 cm
R(Al) =
3.578 cm
= 21 µm
1700
Example 1.28 The range of 5 MeV a’s in air at NTP is 3.8 cm. Estimate the range
of 10 MeV a’s using Geiger-Nuttal law.
1.5 Theory of Ionization
57
3
Solution According to Geiger’s rule, R ∝ v 3 , or R ∝ E 2
3/2
√
E2 3/2
10
=
=2 2
E1
5
√
√
R2 = 2 2R1 = (2 2)(3.8) = 10.75 cm
R2
=
R1
Example 1.29 Mean ranges of a particles in air under standard conditions is defined
by the formula R (cm) = 0.98 × 10−27 v03 , where v0 (cm/s) is the initial velocity of
an alpha particle. Using this formula, find an α-particle with initial kinetic energy
7.0 MeV (a) its mean range (b) the average number of ion pairs formed by the given
a-particle over the whole path as well as over its first half, assuming the ion pair
formation energy to be equal to 34 eV.
Solution
(a)
2T
2T
2×7
v0 =
=c
= 0.061c
=c
2
m
3726
mc
3
R = 0.98 × 10−27 × 3 × 1010 × 0.061 = 6 cm
6
5
(b) (i) Total number of ion pairs = 7×10
34 = 2.06 × 10
3
−27
(ii) For R = 3 cm range, 3 = 0.98 × 10 v0 , or v0 = 1.45 × 109 cm/s. Corresponding energy at mid path is
1
1
1
E = Mv 2 = Mc2 (v/c)2 = × 3726 × (0.048)2 = 4.39 MeV
2
2
2
Energy lost in the first half of the path, E = 7.0–4.39 = 2.61 MeV.
6
4
Number of ion pairs over the first half of the path = 2.61
34 × 10 = 7.67 × 10 .
Example 1.30 Assuming that 14 C and 14 N nuclei are both accelerated to an energy
of 40 MeV and are then allowed to pass through a thin foil. If the 14 C nuclei lose
2 MeV, how much energy will the 14 N nuclei lose?
Solution
−dE
z2
∝ 2
dx
v
As
M
E
or
∝ Z2
M
E
is the same for the nuclei
(−dE/dx)N =
2
zN
zC2
(−dE/dx)C =
72 × 2
= 2.72 MeV
62
58
1
Passage of Charged Particles Through Matter
Example 1.31 Protons and deuterons have the same kinetic energy when they enter
a thin sheet of material. How are their energy losses related?
Solution
−
z2
dE
∝ 2
dx
v
or
∝
M
E
as both P and d have the same z. Also both have same energy E. Therefore,
dE
(− dE
dx )d = 2( dx )p .
Example 1.32 If protons and deuterons lose the same amount of energy when they
enter a thin sheet of material, how are their energies related?
Solution
dE
M
−
∝
dx
E
Mp
Md
=
Ep
Ed
Ed =
Md
2
Ep = Ep = 2Ep
Mp
1
1.5.3 Energy Loss to Electrons and Nuclei
For fast charged particles the energy loss results more from electron collisions than
nuclear collisions. The latter affect stopping mainly for relatively low velocities and
large charges of incident particles. For helium ions of energy larger than 0.5 MeV,
even in heavy materials like silver and gold, the nuclear collisions do not account
for more than 0.5 per cent of the total energy losses. For heavy ions with relatively
low velocities, the contribution of nuclear collisions becomes increasingly important
with charge. However, in this case too the collisions with electrons is the dominant
process for the energy loss. Thus, for example, in the case of quadruply ionized
carbon and oxygen ions in metals, nuclear collisions contribute only to the extent of
a few per cent of the energy loss.
1.5.4 Energy Loss of Heavy Fragments
Heavy ions such as 12 C, 16 O, 40 A, 85 Kr are slowed down predominantly by ionization in much the same way as alpha particles. The only difference is that z is
1.5 Theory of Ionization
59
replaced by zeff = f (β)z, where f (β) is an increasing function of velocity reaching its limiting value of 1 for β = 2z/137. At very low incident particle velocities various complicated effects enter the energy loss mechanism. When the velocity approaches that of K-shell electron, energy transfer to K-electrons becomes
difficult. The energy at which the energy loss attains maximum value is given
by E(max) ∼
= 12 Mc2 (1/137)2 Z 2/3 , where M is the mass of the incident particle,
Z being the atomic number of the target. At velocities (v) less than Bohr’s orbital velocity (u) for K-electron, the incident particle tends to capture an electron
(s) from the atom, resulting in the decrease of the effective charge of the incident
particle. This is called ‘pick-up’ process. It may also lose the captured electron.
The pick-up process becomes a highly probable process for velocity v ≈ u, where
u = zc/137 = 0.22 × 109 cm/s for protons (Ep = 25 keV) and = 0.44 × 109 cm/s
for alphas (Eα = 400 keV). Towards the end of the range, as the velocity decreases,
the stopping power increases reaching the maximum value for β = 0.037 for carbon
and 0.059 for argon-40 ions, which correspond to 8 and 65 MeV energy respectively. At lower energy the stopping power decreases as the ions are further slowed
down, since the decrease of nuclear charge overcompensates the opposing effect of
diminishing velocity. This phenomenon is beautifully demonstrated by the thinning
down of very heavy ion tracks just before they are arrested in photographic emulsions. The extreme case is furnished by the fission fragments. Their effective charge
is large reaching about 20e at the beginning of the range, and nuclear collisions
are an important source of energy loss. If a fragment of atomic number z crosses a
medium of atomic number Z and nuclear mass m2 , the specific energy loss is
−dE
z2 Z 2
∝
dx
m2 v 2
(nuclear)
(1.105)
whereas the loss to electrons is
−
dE
Z
2
∝ zeff
dx
mv 2
(electronic)
(1.106)
Equation (1.105) applies to close nuclear collisions where the entire charges of the
fragments and the target are effective. In the case of electronic collisions, only the
net charge zeff of the fragment is effective, since it carries with it certain number of
electrons, and further the target electrons have unit charge. The factor Z in (1.106)
arises from the presence of Z electrons/nucleus. The two energy losses may be comparable, but only a few nuclear collisions are responsible for the nuclear component
of energy loss whilst in the electronic collisions, the loss is uniformly distributed
along the path. The peculiar branches observed in the cloud chamber photographs
of fission fragments have their origin in nuclear collisions. The concentration of
nuclear energy loss in a limited number of events leads to the enormous spread of
ranges of fission fragments of the same energy, a phenomenon called ‘straggling’.
It is of interest to point out that heavy ions in passing through crystalline solids
lose energy differently depending on the orientation of the trajectory with respect to
the axes of a single crystal. For example, 40 keV 85 Kr ions are found to penetrate
60
1
Passage of Charged Particles Through Matter
the face centred cubic lattice of aluminum crystals for about 4000 Å in the direction
perpendicular to the (101) face but only 1500 Å in the direction perpendicular to the
(111) face. This is because the number of atoms encountered in these two cases is
not same.
1.5.5 Energy Loss of Electrons
It was pointed out that in the case of heavy ions, ionization is the dominant mode of
energy loss. However, for electrons, the energy loss is complicated due to an additional mechanism of loss through radiation, a phenomenon called Bremsstrahlung.
At low energies (E < 2mc2 ) the ionization loss dominates over that due to radiation. The problem of energy loss of electrons by ionization follows similar to that of
heavy ions, but the treatment differs in two important respects. They are the identity
of particles which participate in the collisions, and secondly their reduced mass.
The formula for non-relativistic electrons is:
mv 2 1
1
−dE 4πe4 n
ln
=
−
ln
2
+
(1.107)
dx
2I
2
2
mv 2
Except for small differences in the terms within the square brackets, formula
(1.107) bears a striking resemblance to (1.99). We, therefore, conclude that the nonrelativistic electrons lose their energy by ionization at the same rate as the protons.
The relativistic formula for electrons is
1
dE 4πe4 n
2mc2 3
=
+ ln γ +
−
ln
(1.108)
dx
I
2
16
mc2
and that for protons
2mc2
dE 4πe4 n
ln
=
+ 2 ln γ − 1
−
dx
I
mc2
(1.109)
where γ = 1/ 1 − β 2 is the Lorentz factor. At equal velocities, formulae (1.108)
and (1.109) agree within 10 per cent.
1.6 Delta Rays
1.6.1 Energy Spectrum
In the collision of a charged particle with an atom, one or more electrons are ejected.
The more energetic ones of these are called Delta rays and are responsible for the
secondary ionization, i.e. the production of further ions due to the collision of delta
1.6 Delta Rays
61
rays with the atoms of the medium. In what follows we shall be concerned with
delta rays of energy larger than the ionization potential of the atoms of the medium.
The binding energy of the electron is, therefore, ignored and the collision between
the incident particle and the electrons is considered as approximately elastic. From
(1.27) the kinetic energy of the ejected electron (m2 m1 )
W = 2mv 2 cos2 φ
(1.110)
where m = m2 is the electron mass, φ is the angle of emission of electron, and v is
the velocity of the incident particle. The maximum energy, W (max) = 2mv 2 (nonrelativistically). Now, for the recoil particle (electron) φ = 12 φ ∗ and φ ∗ = π − θ ∗ ,
and so
1
cos2 φ = sin2 θ ∗
2
(1.111)
1
W = 2mv 2 sin2 θ ∗
2
(1.112)
dW = mv 2 sin θ ∗ dθ ∗
(1.113)
But Rutherford’s formula for scattering in the CMS is
z2 e 4
1
dσ
=
σ θ∗ =
dΩ ∗ 4 μ2 v 4 sin4 12 θ ∗
(1.114)
where we have put Z = −1. Since the electron mass is negligible compared to that of
the incident particle, μ ∼
= m. Also, the element of solid angle dΩ ∗ = 2π sin θ ∗ dθ ∗ .
Therefore,
dσ =
2π sin θ ∗ dθ ∗ z2 e4
4m2 v 4 sin4 12 θ ∗
(1.115)
Using (1.112) and (1.113) in (1.115)
dσ
2πz2 e4
=
dW
mv 2 W 2
(differential energy spectrum)
(1.116)
This gives us the cross-section for finding the delta rays of energy W per unit of
energy interval.
1.6.2 Angular Distribution
Using the relations (1.111) and (1.115) and the expression for the element of solid
angle in the lab system dΩ = 2π sin φdφ, we obtain the differential cross-section
62
1
Passage of Charged Particles Through Matter
for the delta rays in the LS:
σ (φ) =
z2 e 4
m2 v 4 cos3 φ
(1.117)
where we have used the relations φ = 12 φ ∗ = 12 π − 12 θ ∗ . It follows that most of the
delta rays are emitted at large angles with correspondingly small energy. Note that
φ(max) = 90◦ for which W = 0. The fact that the delta rays can be emitted only in
the forward hemisphere implies that one can find the direction of the primary.
1.6.3 Delta Ray Density
For a 5 MeV proton W (max) = 2mv 2 = 4T m/M = 4 × 0.51 × 5/940 MeV =
10.85 keV. From (1.116), it is evident that the number of delta rays per cm of path is
inversely proportional to the primary energy; also it is greater for heavy primaries.
The observation of delta ray per cm density is very useful in establishing the charge
of heavy nuclei in cosmic radiation. The total number of δ-rays/cm with energy
> W1 , is given by integrating (1.116) between the limits 2mv 2 corresponding to the
maximum energy of delta rays and some arbitrarily lower value W1 , and multiplying the result by N , the number of electrons per cm3 . This follows from the fact that
n(T , v), the number of δ-rays ejected in 1 cm = 1/λ = Σ = N σ . We thus have:
2πNe4 z2 1
1
(1.118)
n(T , v) =
−
W1 2mv 2
mv 2
Below the lower limit W1 , the δ-rays are not recorded. Clearly, n(T , v) is an arbitrary quantity as it depends on the choice of W1 . It follows that for particles of
identical velocities but of different charges, n(T , v) varies as z2 and the distributions of the values of n(T , v) along the tracks of the particles would, apart from
statistical
√ fluctuations, be similar in form. It is also seen that at a velocity less than
vc = (w1 /2m) the primary would not produce δ-rays with energy > w1 . Above
the critical value, the density would increase at a rate which depends on the variation
of the
√ velocity of the particle along the track. The maximum value is attained for
v = (w1 /m) which is simply obtained by maximizing n with respect to v. After
this, it varies approximately as 1/v 2 , as the second term in the brackets becomes
practically constant. The resulting distribution would thus increase to a maximum
and then slowly decrease (Fig. 1.36). The maximum value n(max) for a given particle of z2 may be compared with that obtained from similar observations on the
tracks of particles of known charge z1 . Thus, the unknown charge z2 may easily be
obtained from the following condition:
n2 /n1 = z22 /z12
(1.119)
It may be pointed out that this condition is also fulfilled for relativistic particles. It is
also possible to determine the mass of the primary particle by measuring the emission angle and the energy of the delta ray caused by the particle whose momentum
1.7 Straggling
63
Fig. 1.36 Variation of δ-ray
density with range for nuclei
of charges 2 to 26
p = Mv is known. We can rewrite (1.110)
2mp 2
cos2 φ
(1.120)
M2
From the measurement of W , φ and p, the mass M of the primary particle can
be deduced. This method is specially suited when the conventional methods do not
permit the particles to be identified. For example, in bubble chambers, this method
is commonly employed for the estimation of contamination of pions or muons in
kaon or antiproton beams.
W = 2mv 2 cos2 φ =
1.7 Straggling
1.7.1 Theory
Identical charged particles, having the same initial velocity, do not have exactly the
same ranges. In other words, for a given energy loss the path length fluctuates. This
phenomenon is called Range straggling. Also, for a given path length the ionization loss and therefore the energy loss fluctuate. This is called Energy straggling.
There is an intimate relation between the two. The observed ranges of individual
particles from any mono-energetic source will show a substantially normal distribution about the mean range. The standard deviation of this distribution is of the
order of 1 per cent for a few MeV alphas in any absorber. The distribution is due to
the statistical fluctuations in the individual collisions between the charged particle
and atomic electrons, which are finite in number. The nuclear collisions, fewer in
number, which may cause substantial loss of energy specially towards the end of
the ranges, contribute to the short range tail of the distribution. For small energies,
however, this will be a small contribution and the distribution may be taken as approximately symmetrical. The harder collisions account for most of the straggling
and because very hard collisions are few in number, the actual distribution is some
what asymmetric, with a longer tail in the direction of short ranges and with a mean
range slightly shorter than the modal range.
64
1
Passage of Charged Particles Through Matter
1.7.2 Energy Straggling
The energy straggling is produced when an initially mono-energetic beam of particles traverses a given thickness of the absorber. Let Ax be the number of collisions
per unit path length in which an energy between W , and W + dW is transfered.
Then, from (1.116) we have
Ax =
2πNz2 e4 W
mv 2 Wx2
(1.121)
where N is the number of electrons/cm3 . The energy transfer in a distance r is
given by
E =
Ax Wx r
(1.122)
x
E/
r =
Ax Wx
(1.123)
x
When the number of collisions is large, we may use integration rather than summation
dE 2πNz2 e4 W (max) dW
=
(1.124)
dr
mv 2
W (min) Wx
The statistical fluctuations in energy loss E arise from fluctuations about the avassume that the collisions are randomly diserage number of collisions Ax r. We√
Ax r. The S.D. of the energy loss is then
tributed
and
that
the
S.D.
is
given
by
√
Wx Ax r. The variance for all types of collisions is then given by the summation
of the individual variances
2πNz2 e4 r W (max)
2
2
Wx Ax =
dW
σ = r
mv 2
W (min)
x
=
2πNZ 2 e4 r W (max) − W (min)
mv 2
where we have replaced the summation by integration. Since W (min) W (max) =
2mv 2
σ 2 = 4πNz2 e4 r
(1.125)
If it is assumed that the actual energy loss has a Gaussian distribution around the
average value E0 , the use of expression (1.125) for the S.D. in energy loss leads to
(E − E0 )2
dE
P (E)dE = √
exp −
(1.126)
8πNz2 e4 t
8π 2 z2 e4 N t
where t is the absorber thickness.
1.7 Straggling
65
Fig. 1.37 Energy distribution of an ‘unobstructed’ electron beam and the calculated and experimental distributions of electrons that have passed through 0.86 g cm−2 of aluminum. (1) Landau
theory without density correction; (2) Landau theory with Fermi density correction; (3) experiment; (4) incident beam [1]
In the case of fission fragments large energy losses in individual nuclear collisions give rise to a tail on the side of higher energy losses of the distribution. The
straggling effects are much more important for electrons than for heavy particles,
because an electron may lose even half its energy in a single elastic collision, where
as a heavy particle may lose only a fraction of its energy. Radiation losses add further
to the electron straggling. Thus electron straggling reaches values of the order of 0.2
of the total energy loss. Figure 1.37 shows the energy distribution of electrons before
they have entered the absorber and after they have traversed 0.86 g/cm2 thickness
of aluminium.
1.7.3 Range Straggling
The fluctuations in range and energy loss are related. Denoting the S.D. of energy
and range by σE and σR respectively, we can use the formula for the propagation of
errors and write:
σR2 = (dR/dE)2 σE2
(1.127)
σR2 = (dR/dE)2 4πNz2 e4 dR
(1.128)
Using (1.125), we get
Writing dR = (dR/dE)dE, we get the result:
E0 dE −3
σR2 = 4πNz2 e4
dE
dR
0
(1.129)
This relation is not applicable to heavy ions and fission fragments that undergo
excessive straggling owing to the occurrence of single nuclear collisions. Assuming
66
1
Passage of Charged Particles Through Matter
Fig. 1.38 Measured ranges
of muons from π –μ decay in
emulsions of standard
composition
that the ranges of individual particles are distributed about the mean range in a
Gaussian way, the probability that the individual range falls between R and R + dR
is
dR
P (R)dR = √ exp − (R − R)2 /2σR2
(1.130)
σR 2π
For α particles from Polonium, E0 = 5.3 MeV, R = 3.84 cm in air, the corresponding σR = 0.036 cm and the ratio σR /R = 0.9 %. Figure 1.38 shows the histogram of
ranges of μ mesons produced in the decay of π + mesons at rest. Since the π mesons
decay by a two-body process, μ+ is produced with unique energy (4.27 MeV). The
mean range in photographic emulsions is found to be 600 µm. The S.D. of the range
distribution is found to be, σR = 2.7 µm; this gives σR /R = 0.045, or 4.5 percent.
1.7.3.1 The Range Straggling Parameter
This is related to S.D. by
α0 =
√
2σR
(1.131)
Several common types of particle detectors measure the integrated number of particles. The particles that are still present in the collimated beam having ranges equal
or greater than R is given by
n = n0 −
R
−∞
dn
where dn/n0 is given by the normal distribution
dn
1
= √ exp −(R − R)2 /α 2 dR
n0
α π
(1.132)
dn is the actual range between R and R + dR, n0 is the total number of particles
initially present, and α is the half width of the range distribution at 1/e of the max-
1.7 Straggling
67
Fig. 1.39 The extrapolated
number-distance range Rn
exceeds the mean range R by
0.886α, where α is the
range-straggling parameter
imum. Although the normal distribution is non-integrable, its value can be found
from standard tables.
The number-distance curve, n/n0 against R,√is indicated in Fig. 1.39. Its slope
(dn/n0 )/dR at the mean range R = R is 1/α π . As the central portion of the
number-distance curve is approximately linear, it can be extrapolated to cut the
range axis at R = Rn . This is called extrapolated range. From Fig. 1.39, we find
the relation between Rn and R
1
2
Rn − R
1
= √
α π
(1.133)
whence the mean range, in term of the measured extrapolated range Rn and straggling parameter a, is
R = Rn −
1√
π a = Rn − 0.886α
2
(1.134)
1.7.3.2 Deduction of Ranges Parameter
For particles of charge ze and mass M but the same initial velocity v0 as alphaparticles
σR = 4πNz2 e4
0
E0 dE −3
dR
1/2
dE
dE 2πz2 e4 N w(max) dw
=
dR
w
mv 2
w(min)
√
M
dE = d Mv 2 = Mvdv and σR =
f (v0 , I )
N z2
dE
= z2 f (v0 )N,
dR
R=
Mf (v0 , I )
N z2
since
(1.135)
68
1
Passage of Charged Particles Through Matter
σR
1
= √ f (v0 , I )
R
M
(1.136)
where f , f , f are complicated functions of the initial velocity v0 , mean excitation
and ionization potential I . The function f and hence σR /R are independent of N
and of z but is found to decrease slowly with increasing I .
For particles of mass M having the same initial velocity v0 as the α particles,
(α0 /R)M
4
=
(1.137)
(α0 /R)α particle
M
It follows that protons will have about twice the range straggling parameter of αparticles which have the same initial velocity and hence about the same range.
1.8 Cerenkov Radiation
Electromagnetic radiation is emitted when a charged particle passes through a
medium in which its velocity v = βc exceeds the phase velocity c/μ, where μ is the
refractive index of the medium. This observation was discovered by Cerenkov and
was explained theoretically by Frank and Tamm. The effect was first observed in the
experiments of Cerenkov who was investigating the glow in pure liquids caused by
γ rays from radium. Vavilov and Cerenkov showed that the radiation is not due to
luminescence (emission from excited atoms and molecules of the medium) but due
to the passage of knock-on electrons produced in Compton scattering of γ rays. The
radiation is instantaneous and possesses a sharply pronounced spatial symmetry.
When relativistic charged particles are incident on a transparent dielectric, the
velocity of the particle is substantially unchanged except for the ionization and radiation losses. On the other hand, the electric field due to the charge of the particle
and the magnetic field produced by the moving charge are propagated through the
medium with velocity of only c/μ. The resulting electromagnetic radiation is cancelled in all directions if βμ < 1; however, if βμ > 1, constructive interference can
take place in one direction defined by angle θ (Fig. 1.40). When βc > c/μ, i.e. the
particle velocity exceeds the velocity of light in the medium it is as if the particle
runs away from its own slower electromagnetic field, resulting in the emission of
all frequencies for which βμ > 1. The resulting radiation called Cerenkov radiation is emitted on a conical surface BDA of half angle a0 . Figure 1.40 gives the
Huyghens’ construction for the electromagnetic waves emitted by the particle along
its path. The particle is at A at t = 0; and at a later time it moves on to D such that
AD = βct. The front of the electromagnetic wave lies on the surface of the cone of
half angle α. Consequently the corresponding rays of light make an angle θ with the
path of the particle. The axis of the cone coincides with the direction of the incident
particle and the half angle of the cone is determined by:
sin α = cos θ =
(c/μ)t
1
=
βct
βμ
(1.138)
1.8 Cerenkov Radiation
69
Fig. 1.40 Huyghens’
construction for
electromagnetic waves
emitted by a moving charged
particle
This follows from the condition that the optical difference in the path of the waves
emitted by the moving particle at various points of its trajectory is equal to zero.
The light is polarized with its electric vector in the plane of the conical surface
and radially directed along DB. The conical distribution of the Cerenkov radiation
has a natural half width of the order of a few degrees. This is attributed to the occurrence of successive changes in particle velocity when photons are emitted. The
phenomenon is analogous to the V-shaped shock wave observed in acoustics when
a projectile or an aeroplane travels with supersonic velocity. Apart from (1.138),
there are two other conditions that must be fulfilled to achieve coherence. These
are (i) pathlength of the particle in the medium must be large compared with the
wavelength of the radiation, otherwise diffraction effects become dominant and (ii)
velocity of the particle must remain constant during its passage through the medium.
For a medium of a given refractive index μ, there is a threshold velocity
β(min) = 1/μ, below which no radiation is emitted. At this critical velocity, the
direction of radiation coincides with that of the particle. For glass (μ = 1.5),
β(min) = 0.667, corresponding to 200 keV electrons or 320 MeV protons. As the
refractive index decreases, the threshold velocity increases. For an ultra-relativistic
particle, for which β 1, there is a maximum angle of emission given by θ (max) =
cos−1 (1/μ).
Fermi showed that Cerenkov radiation results from small energy transfers to distant atoms due to the fast moving charged particles which is subsequently emitted as
a coherent radiation. Thus the emission of Cerenkov radiation is a particular form of
energy loss in extremely soft collisions. The classical theory of Cerenkov effect is
originally due to Frank and Tamm and is justified by the quantum theory. Since the
radiation in question is believed to be the result of the interaction with the medium
as a whole and not due to the interaction of particles with individual atoms, the
medium is considered as continuous and is characterized by the macroscopic parameter, the dielectric constant or by the refractive index. It is shown that the rate of
energy loss per unit path length is given by:
−
dE 4π 2 z2 e2
=
dx
c2
1−
βμ>1
1
νdν ergs/cm
β 2 μ2
(1.139)
where ze is the charge of the particle and ν is the frequency of the emitted radiation. The integration is to be carried over all frequencies for which βμ > 1. For
glass or Lucite, the energy loss by Cerenkov radiation is of the order of 1 keV/cm,
a value which is much less than that incurred in ionization or radiation. Nonetheless
70
1
Passage of Charged Particles Through Matter
the radiation is readily detected as a large number of photons are produced in the
visible region. Formula (1.139) shows that the Cerenkov radiation is independent
of the rest mass of the moving particle and depends only on the particle’s charge
and velocity, apart from the refractive index of the medium. The mean number of
photons of frequency ν and ν + dν in the visible region per cm is calculated in Example 1.35, under the assumption that μ is independent of ν in the considered range
of frequencies. Hence
1
2πz2 dν sin2 θ
4π 2 z2 e2
1
−
dν
=
N(ν)dν =
137c
hc2
μ2 β 2
(1.140)
The radiation has continuous spectrum, with components of all frequencies for
which the refractive indices are higher than 1/β. Equation (1.139) shows via the
term νdν which is proportional to dλ/λ3 that the energy per wavelength interval dλ
is proportional to 1/λ3 . Also, (1.140) shows through the term dυ that the number
of quanta per cm per wavelength interval is proportional to 1/λ2 . It follows that
shorter wavelengths are preferred and the Cerenkov radiation appears visually as
bluish white.
The density effect is closely connected with the phenomenon of Cerenkov effect.
It was first pointed out by Bohr that the intricate relationship between the density
effect and the Cerenkov effect is such that the entire contribution to the most probable energy loss from the minimum out to the beginning of the Fermi plateau in the
ionization curve is due to Cerenkov effect.
Example 1.33 Pions and muons each of 160 MeV/c momentum pass through a
transparent material. Find the range of the index of refraction of this material over
which the muons alone give Cerenkov light. Assume mπ c2 = 140 MeV, mμ c2 =
106 MeV.
Solution Momentum, p = mβγ c. Therefore, √ β
1−β 2
Pions:
cp
mc2
β
cp
160 8
= =
=
mπ c2 140 7
1 − β2
βπ = 0.7525;
Muons:
=
μπ =
1
1
= 1.33
=
βπ
0.7525
cp
β
160
=
=
mμ c2 106
1 − β2
βμ = 0.8336;
μμ =
1
1
= 1.2
=
βμ 0.8336
Therefore, the range of the index of refraction of the material over which the muons
alone give Cerenkov light is 1.2–1.33.
1.8 Cerenkov Radiation
71
Example 1.34 A beam of protons moves through a material whose refractive index
is 1.6. Cerenkov light is emitted at an angle of 15° to the beam. Find the kinetic
energy of the proton in MeV.
Solution
β=
1
1
=
= 0.647
μ cos θ
1.6 cos 15◦
γ=
1
1 − β2
=
1
1 − (0.647)2
= 1.31
K.E. = (γ − 1)mc2 = (1.31 − 1) × 938 = 292 MeV
Example 1.35 The rate of loss of energy by production of Cerenkov radiation is
given by the relation
−dW/dl =
z2 e 2
c2
1−
1
ωdω erg cm−1
β 2 μ2
where βc is the velocity of the of charge ze, μ is the refractive index of the medium
and ω/2π is the frequency of radiation. Estimate the number of photons emitted
in the visible region, per cm of track, by a particle having β = 0.8 passing through
glass (μ = 1.5). The fine structure constant α = e2 /c = 1/137.
Solution For electron, z = −1 and since ω = 2πν, the given expression becomes
upon integration between the frequencies ν1 and ν2
−dW/dl =
2
(ν2 − ν12 )
1
4π 2 e2
1
−
2
c2
β 2 μ2
where we have assumed that μ is independent of ν.
Calling the average photon energy as hν = 12 h(ν1 + ν2 ), the average number of
quanta emitted per cm is
1
4π 2 e2
1 −dνc
1
−
(ν2 − ν1 )
=
hν
dl
hc2
β 2 μ2
1
1
1
2π
1− 2 2
−
=
137
λ2 λ1
β μ
N=
where λ = c/ν is the vacuum wavelength and μ is the average refractive index
over the wavelength interval from λ2 = 4000 Å to λ1 = 8000 Å and βμ = 0.8 ×
1.5 = 1.2.
2π
1
1
1
N=
= 175 photons
1−
−
137
1.22
4000 × 10−8 8000 × 10−8
72
1
Passage of Charged Particles Through Matter
1.9 Identification of Charged Particles
In numerous investigations in nuclear physics and particle physics it is necessary
to determine the nature and energy of charged particles. In order to identify a particle, its charge and mass must be determined. The charge may be determined by
δ-ray density or width measurements in photographic emulsions or pulse height in
scintillation counters or proportional counters or solid state detectors.
Assuming that the charge is known, the mass of the particle can be determined
from the simultaneous measurements of at least two dynamical quantities such
as momentum and velocity, momentum times velocity and −dE/dX, −dE/dX
and E.
1.9.1 (a) Momentum and Velocity
Momentum can be determined from curvature measurement in a cloud chamber or
in a bubble chamber with low Z liquid with known magnetic field or in photographic
emulsions with pulsed magnetic field.
Velocity may be estimated by the estimation of ionization through drop density
in a cloud chamber, bubble density in a bubble chamber, grain density, blob density
or mean gap length in photographic emulsions or by Cerenkov counters or time-offlight method.
1.9.2 (b) Momentum Times Velocity (pβ) and Velocity
The product pβ is determined from the mean scattering angle in emulsions for energetic particles. The velocity is measured as in Sect. 1.9.1.
1.9.3 Energy and Velocity
Energy may be determined from range measurements for low energy particles and
velocity as in Sect. 1.9.1.
1.9.4 Simultaneous Measurement of dE/dx and E
This method is widely used in the study of nuclear reactions using solid state detectors, since for non-relativistic particles the product EdE/dx is proportional to
z2 M. A simultaneous measurement of E and dE/dx and their product permits the
separation of the particles according to their masses in a wide range of energy variations.
1.10
Bremsstrahlung
73
1.9.5 Energy and Emission Angle
Energy and emission angle measurement of knock-on electron (δ-ray) together with
the momentum measurement of the beam particles.
This method is very useful in case the conventional methods are not available.
The method is explained in Sect. 1.6.3.
1.10 Bremsstrahlung
In their passage through matter, electrons lose energy in two ways (i) ionization
(which was referred to in Sect. 1.5.1) and (ii) radiation or Bremsstrahlung. The electrons undergo radiative collisions mainly with the atomic nuclei of the medium. In
the vicinity of the nucleus of charge Ze, the incident particle of charge ze and mass
m undergoes acceleration which is proportional to zZ/m. According to electrodynamics, a charged particle undergoing acceleration emits radiation. This is called
Bremsstrahlung or braking radiation whose spectrum has the form dE/E where E
is the photon energy. The photon energy spectrum extends from low energy to the
maximum value equal to the particle energy, with the low energy photons being
preferably emitted (see Fig. 1.41 for typical energy spectrum). The radiation intensity is proportional to z2 Z 2 /m2 . This then means that under identical conditions,
radiation losses are 3 × 106 times as much for electron as for a proton. The total
average energy loss per path length dx integrated over all frequencies is given by
−(dE)rad =
4Z(Z + 1)
183
N Ere2 ln 1/3 dx
137
Z
(1.141)
where N is the number of nuclei per cm3 , E is the energy of electron and re =
e2 /mc2 is the classical electron radius.
Since an electron may lose appreciable energy in a single collision, the actual
energy loss may vary significantly from the average value given by (1.141). This also
implies that the range straggling of electrons would be so great that the definition of
mean range would hardly be meaningful. If we define the radiation length X0 by
1
4Z(Z + 1) 2
183
re N ln 1/3
=
X0
137
Z
(1.142)
dE
E
=−
dx
X0
(1.143)
we can write from (1.141)
Integrating (1.143), we find the average energy of a beam of electrons of initial
energy E0 after traversing a thickness x of medium by the expression
E = E0 exp(−x/X0 )
(1.144)
74
1
Passage of Charged Particles Through Matter
Fig. 1.41 Energy distribution of the radiation emitted by an electron. Ordinate intensity of radiation (quantum energy times number of quanta) per unit frequency interval. Abscissa, energy of
emitted quantum as a fraction of the energy of the emitting electron. The numbers on the curves
indicate the energy of the electron in units of mc2 . Solid curves for lead, including effect of screening. Dotted curves are without screening, valid for all Z [3]
For x = X0 , E = E0 /e, where e is the exponential. This suggests that the radiation
length X0 may be simply defined as that thickness of the medium which reduces the
beam energy by a factor of e. Since the thickness x can be measured in cm or g/cm2
(which is obtained by multiplying the thickness in cm by the density of the medium)
X0 , is expressed in corresponding units. At low electron energies (E mc2 ), the
electrons lose their energy predominantly through excitation and ionization, and
radiation loss is unimportant. The energy loss by ionization and excitation is proportional to Z and is practically constant at high energies as it increases only logarithmically with energy. On the other hand, radiation losses are proportional to Z 2
and increase linearly with energy. Thus, the radiation loss predominates at high energies. It is apparent that at some energy Ec , called the critical energy, Erad = Eion .
It can be shown that roughly
(dE/dx)rad
EX
=
(dE/dx)ion
600
(1.145)
so that Ec (in MeV) = 600/x. The radiation lengths X0 and the critical energy Ec ,
for some of the materials are shown in Table 1.1. Observe that X0 , decreases rapidly
with increasing Z.
1.11 Questions
1.1 Why in Rutherford scattering the presence of orbital electrons in the target
atom is ignored?
1.2 The total cross-section for Rutherford scattering is infinite. What is the physical
reason?
1.11
Questions
Table 1.1 Radiation lengths
and critical energy in
different elements
75
Element
Z
X0 (g cm−2 )
Ec (MeV)
Hydrogen
1
58
340
Carbon
6
42.5
103
Air
7.2
36.5
83
Aluminium
11
23.9
47
Iron
26
13.8
24
Lead
82
5.8
6.9
1.3 Why in the famous a-scattering experiment thin foils were used for the target?
1.4 If the incident electron enters the nucleus, would the Coulomb’s inverse square
law between the charges be still valid? If not, how would it be modified for a nucleus
in which the charge is uniformly distributed?
1.5 Why does the ionization fall off for very low particle velocity?
1.6 The inverse square velocity law for ionization would suggest that the rate of
energy loss is greater at low speeds, since the time spent by the incident particle in
the vicinity of the electron is longer. Is this reasonable? In the same manner would
a slow moving heavenly object raise larger tides on approaching close to the earth
compared to a fast moving one?
1.7 What is the physical origin of the rise in the −dE/dx curve beyond the minimum?
1.8 At relativistic velocities, the −dE/dx curve saturates to a plateau. What is the
origin of the plateau?
1.9 In the cloud chamber studies of ionization, the plateau-to-trough ratio for the
−dE/dx curve might be as large as 1.5, but in photographic emulsions it is no more
than 1.1. Explain.
1.10 How does the percentage straggling compare for 3 H and 3 He nuclei of the
same initial velocity?
1.11 A cloud chamber photograph shows an alpha track which after certain distance gets thinned down and then disappears. It again re-appears before it stops.
What is happening?
1.12 The range of a proton of few MeV is a measure of its initial energy. The energy
thus estimated would be close to the actual value within few per cent. However, in
the case of electrons of similar energy, the energy thus estimated can hardly be
reliable. Explain.
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Passage of Charged Particles Through Matter
1.13 The tracks of fission fragments often leave peculiar branches before coming
to a rest. Explain.
1.14 A water cooled nuclear reactor appears bluish. What could be the origin of
this colour?
1.15 A charged particle moves swiftly with uniform velocity in a vacuum. Would
it radiate?
1.16 What is the dominant mechanism for energy loss for electrons of energy
(a) <1 MeV, (b) 200–500 MeV?
1.12 Problems
1.1 Show that in an elastic collision, the ratio of the kinetic energy K /K can be
expressed through α = M/m and y = cos θ as
K
= (1 + α)−2 2y 2 + α 2 − 1 + 2y a 2 + y 2 − 1
K
1.2 A body of mass M rests on a smooth table. Another of mass m moving with
a velocity u collides with it. Both are perfectly elastic and smooth and no rotations
are set up by the collision. The body M is driven in a direction at an angle φ to the
previous line of motion of the body m. Show that its velocity is
2mu
cos φ
M +m
1.3 A nucleus A of mass 2m moving with velocity u collides inelastically with
the nucleus B of mass 10m. After the collision, the nucleus A travels at 90° with
the incident direction, while B proceeds at an angle 37° with the incident direction.
(a) Find the speeds of A and B after the collision. (b) What fraction of the initial
kinetic energy is gained or lost due to the collision?
[Ans. (a) vA = 3u/4; vB = u4 ; (b) 1/8]
1.4 A beam of alphas gets scattered from a hydrogen target. What is the maximum
angle of scattering?
[Ans. approximately 15°]
1.5 An alpha particle fired into a cloud chamber undergoes an elastic collision with
a nucleus of the gas used to fill the chamber. The collision is recorded photographically as a forked track. Measurements from the photograph show that the collision
deviated the alpha-particle at 60° and that the struck nucleus recoiled at an angle of
30° with the direction of motion of the incident alpha-particle. Assuming that the
struck nucleus is initially at rest, calculate:
1.12
Problems
77
(a) The mass number of the gas used to fill the chamber.
(b) The ratio of the velocity of projection of the struck nucleus to the velocity of
the incident alpha particle.
√
[Ans. (a) 4; (b) 3]
1.6 A particle of mass m makes an elastic collision with a proton, initially at rest.
The proton is projected at an angle 22.1° whilst the incident particle is scattered
through an angle 5.6° with the incident direction. Estimate m in atomic mass units.
[Ans. 7.8]
1.7 Consider an elastic collision between an incident particle having mass m and a
target particle of mass M such that m > M. Show that the largest possible scattering
angle θ (max) in the lab system is given by: sin θ (max) = M/m; and that this corresponds to C-system angle cos θ ∗ (max) = −M/m.
Also show that the maximum
√
recoil angle φ(max) is given by sin φ(max) = (m − M)/2m. Calculate the angle
θ (max) + φ(max) for elastic collisions between the incident deuterons and the target
protons.
[Ans. 60°]
1.8 A billiard ball moving at a speed of 2.5 m/s makes a glancing collision with
another identical ball initially at rest. After the collision, one ball is observed to
move with a speed 2 m/s at an angle 37° with the original direction of motion. Find
the speed of the other ball and the angle at which it moves. What is the nature of the
collision?
[Ans. 1.5 m/s, 53°, elastic]
1.9 If a particle of mass m moving with kinetic energy K0 makes elastic collision
with a target particle of mass M initially at rest, such that the scattered particle is
deflected at an angle θ in the lab system and has θ ∗ in the centre of mass system and
has a kinetic energy K in the lab system, show that:
2
K
1
m cos θ + M cos θ ∗ − θ
=
2
K0 (M + m)
1.10 A particle of mass m and initially of velocity u makes an elastic collision with
a particle of mass M initially
√ at rest. After the collision m is deflected through lab
angle 90° with speed u/ 3. The particle M recoils with speed v at a lab angle φ
with the incident direction.
Find (a) M/m, (b) v/u, (c) φ, (d) θ ∗ , (e) φ ∗ .
√
[Ans. (a) 2, (b) 1/ 3, (c) 30°, (d) 120°, (e) 60°]
1.11 A deuteron of velocity u strikes another deuteron (twice the mass of proton) initially at rest. As a result of the collision, a proton is produced which moves
off at 45° with respect to the direction of incidence. The other product of this rearrangement collision is triton (three times the mass of proton). Assuming that this
78
1
Passage of Charged Particles Through Matter
collision may be approximated to an elastic collision, calculate the speed and direction of triton in the lab and CM system.
√
[Ans. 0.48 u, 34° in the lab system and u/2 3, 111° in the CM system]
1.12 An α-particle from a radioactive source collides with a stationary proton and
continues with a deflection of 13.9°. Find the direction in which the proton moves.
[Ans. 30°]
1.13 When α-particles of kinetic energy 30 MeV pass through a gas, they are
found to be elastically scattered at angles up to 30° but not beyond. Explain this,
and identify the gas. In what way, if any, does the limiting angle vary with energy?
[Ans. Deuterium, does not vary]
1.14 A perfectly smooth sphere of mass m, moving with velocity v collides elastically with a similar but initially stationary sphere of mass m2 (m1 > m2 ) and is
deflected through an angle θL . Describe how this collision would appear in the centre of mass frame of reference and show that the relation between θL and the angle
of deflection θM , in the centre of mass frame is
tan θL =
sin θM
[M1 /M2 + cos θM ]
Also show that θL cannot be greater than about 19.5° if M1 /M2 = 3.
1.15 Show that the maximum velocity that can be imparted to a proton at rest by a
non-relativistic alpha particles is 1.6 times the velocity of the incident alpha particle.
1.16 Show that for low energy p–p scattering σ (θ ) = 4σ (θ ∗ ) where the differential cross-sections σ (θ ) and σ (θ ∗ ) refer to the Lab and CMS, respectively.
1.17 (a) Compute the distance of closest approach in collisions between α-particles
of energy 8.9 MeV and nuclei of 208
82 Pb.
(b) How is this distance related to the radius of lead nucleus?
(c) What is the deflection of the α-particle when the impact parameter is equal to
this distance?
[Ans. (a) 26.5 fm, (b) 7.7 fm, (c) 53°]
1.18 A beam of α-particles of kinetic energy 4.5 MeV passes through a thin foil
of 94 Be. The number of alphas scattered between 60° and 90° and between 90° and
120° is measured. What would be the ratio of these numbers?
[Ans. 3]
1.19 If the probability of α-particles of energy 8 MeV to be scattered through an
angle greater than θ on passing through a thin foil is 10−3 what is it for 4 MeV
protons passing through the same foil?
[Ans. 10−3 ]
1.12
Problems
79
1.20 What α-particle energy would be necessary in order to explore the field of
force within a radius of 10−12 cm of the centre of nucleus of atomic number 80,
assuming classical mechanics to be adequate?
[Ans. 30 MeV]
1.21 In an elastic collision with a heavy nucleus, when the impact parameter b is
just equal to the collision radius 12 R0 , what is the value of the scattering angle θ ∗ in
the CMS?
[Ans. 90°]
1.22 In the elastic scattering of deuterons of 5.9 MeV from 208
82 Pb, the differential
cross-section is observed to deviate from Rutherford’s classical prediction at 52°.
Use the simplest classical model to calculate the closest distance of approach d
to which this angle of scattering corresponds. You are given that for an angle of
scattering θ , d is given by 12 d0 (1 + cosec 12 θ ), where d0 is the value of d in a headon collision.
[Ans. 32.8 fm]
1.23 20000,1 MeV α-particles are incident normally on a 0.004 mm thick copper
plate. Using the small angle approximation, calculate the number of α-particles scattered in the angular range 5°–10°. Assume the copper nuclei to act as point charges
−3
and neglect nuclear forces. Density of 66.6
29 Cu = 8.9 g cm ; Avagadro’s number
= 6 × 1023 (g molecule), e = 1.6 × 10−19 C; 1 eV = 1.6 × 10−19 J.
[Ans. 7894]
1.24 Given that the angle of scattering is 2 tan−1 (a/2b), where ‘a’ is the least
possible distance of approach, and b is the impact parameter. Calculate what fraction
of a beam of 1.0 MeV deuterons will be scattered through more than 90° by a foil
of thickness 10−5 cm of a metal of density 5 g cm−3 atomic weight 100 and atomic
number 50.
[Ans. 1.22 × 10−5 ]
1.25 Show that the differential cross-section for the recoil nucleus in the lab system
is given by
2
σ (φ) = zZe2 /2T
1
cos3 φ
1.26 An electron of energy 10 keV approaches a bare nucleus (Z = 20) with an
impact parameter corresponding to an orbital angular momentum . Sketch the form
of the potential energy curve for the electron trajectory and calculate the distance
from the nucleus at which this has a minimum (take = 10−34 J s, e = 1.6×10−19 C
and m = 10−30 kg).
[Ans. 0.19 Å]
80
1
Passage of Charged Particles Through Matter
1.27 A beam of protons of 5 MeV kinetic energy traverses a gold foil. One particle
in 5 × 106 is scattered so as to hit a surface 0.5 cm2 in area at a distance 10 cm from
the foil and in a direction making an angle of 60° with the initial direction of the
beam. What is the thickness of the foil?
[Ans. 0.0066 µm]
1.28 A narrow beam of protons with velocity v = 6 × 106 m/s falls normally on a
silver foil of thickness t = 1.0 µm. Find the probability of the protons to be scattered
into the backward hemisphere (θ > 90◦ ).
[Ans. 0.006]
1.29 A narrow beam of alpha particles with K.E. 0.5 MeV falls normally on a
golden foil whose thickness is 1.5 mg/cm2 . The beam intensity is 5 × 105 particles
per sec. Find the number of alpha particles scattered by the foil during the time
interval of 30 minutes into angular interval 59–61°.
[Ans. 1.6 × 106 ]
1.30 A narrow beam of alpha particles falls normally on a silver foil behind which
a counter is set to register the scattered particles. On substitution of platinum foil of
the same mass thickness for the silver foil, the number of alpha particles registered
per unit time increases 1.52 times. Find the atomic number of platinum, assuming
the atomic number of silver and the atomic masses of both platinum and silver to be
known.
[Ans. 78]
1.31 A narrow beam of alpha particles with kinetic energy 1.0 MeV falls normally
on a platinum foil which is 1.0 µm thick. The scattered particles are observed at an
angle of 60° to the incident beam direction by means of a counter with a circular
sensitive area 1.0 cm2 located at a distance 10 cm from the scattering section of
the foil. What fraction of scattered alpha particles enters the counter? Assume the
density of platinum as 21.5 g/cm3 .
[Ans. 3.33 × 10−5 ]
1.32 Singly charged particles of masses m1 and m2 enter a medium with the same
velocity. Show that the ratio of their ranges R1 /R2 = m1 /m2 .
1.33 Show that a deuteron of energy E has twice the range of a protons of energy
E/2.
1.34 If the mean range of 8 MeV proton in a medium in 0.30 mm, calculate the
mean range of 16 MeV deuterons and 32 MeV α-particles.
1.35 An alpha particle moving with velocity 2 × 109 cm/sec, loses energy
0.066 MeV/mm by ionization in air and has range 7.86 cm in air. (a) Find the rate of
References
81
loss of energy per mm in air for proton and deuteron moving with the same initial
velocity as alpha particle. (b) Find the range of proton and deuteron.
[Ans. (a) 0.0165 MeV/mm for both, (b) 7.86 cm, 15.72 cm]
1.36 Estimate by the Bragg-Kleeman rule the mean range of 12 MeV deuterons in
cobalt, if their mean range in air at 15 °C, 760 mm Hg is 93 cm. Assume the density
of cobalt to be 8.6 g/cm3 .
[Ans. 0.0266 cm]
1.37 Show that the range of α-particles and protons of energy 1 to 10 MeV in
aluminium is 1/1600 of the range in air at 15 °C, 760 mm of Hg.
1.38 Show that the straggling of a beam of 4 He is smaller than that of 3 He of equal
range.
1.39 Compute the energy loss and the approximate number of quanta of visible
light (λ = 4000 to 7000 Å) as Cerenkov radiation by a 20 MeV electron in traversing
1 cm of Lucite. Assume the chemical composition of Lucite to be (C5 H8 O2 ), and
the refractive index μ = 1.5.
[Ans. 660 eV/cm, 270 quanta/cm]
1.40 Show that the order of magnitude of the ratio of the rate of loss of kinetic
energy by radiation for a 10 MeV deuteron and a 10 MeV electron passing through
lead is 10−5 .
1.41 Compute the energy loss and the approximate number of quanta of visible
light (λ = 4000 to 7000 Å) as Cerenkov radiation by a 20 MeV electron in traversing
1 cm of Lucite. Assume the chemical composition of Lucite to be (C5 H8 O2 ), and
the refractive index μ = 1.5.
1.42 Extensive air showers in cosmic rays consist of a ‘soft’ component of electrons and photons, and a ‘hard’ component of muons. Suppose at the sea level the
central core of a shower consists of a narrow vertical beam of muons of energy
100 GeV which penetrate the interior of the earth. Assuming that the ionization loss
in rock is constant at 2 MeV g−1 cm2 , and the rock density is 3.0 g cm−3 , find the
depth of the rock through which the muons can penetrate.
[Ans. 160 m]
References
1. E.L. Goldwasser, F.E. Mills, A.O. Hanson, Phys. Rev. 88, 1137 (1952)
2. A.A. Kamal, G.K. Rao, Y.V. Rao, Osmania University
3. Proc. R. Soc. A 146, 96 (1934)
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